Class 11

Samacheer Kalvi 11th Books Solutions Guide

Expert Teachers at SamacheerKalvi.Guru has created Tamilnadu State Board Samacheer Kalvi 11th Books Answers and Solutions Guide Pdf Free Download of Volume 1 and Volume 2 in English Medium and Tamil Medium are part of Samacheer Kalvi Books Solutions. Here we have given TN State Board New Syllabus Samacheer Kalvi 11th Std Guide Pdf of Book Back Questions and Answers, Chapter Wise Important Questions, Study Material, Question Bank, Notes, Formulas.

Students can also read Tamil Nadu 11th Model Question Papers 2020-2021.

Samacheer Kalvi 11th Guide Book Back Answers Solutions Pdf Free Download

TN Samacheer Kalvi 11th Book Back Answers Solutions Guide

We hope the given Tamilnadu State Board Samacheer Kalvi Class 11th Books Volume 1 and Volume 2 Solutions and Answers Guide Pdf Free Download in English Medium and Tamil Medium will help you. If you have any queries regarding TN State Board New Syllabus Samacheer Kalvi 11th Standard Guides Pdf of Book Back Questions and Answers, Chapter Wise Important Questions, Study Material, Question Bank, Notes, Formulas, drop a comment below and we will get back to you at the earliest.

Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

Expanding Binomials Calculator online with solution and steps.

Question 1.
Expand the following in ascending powers of x and find the condition on x for which the binomial expansion is valid.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 1
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 2
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 3

Question 2.
Find \(\sqrt[3]{1001}\) approximately (two decimal places).
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 4

Question 3.
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 5
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 6

Question 4.
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 7
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 8
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 9

Question 5.
Write the first 6 terms of the exponential series
(i) e5x
(ii) e-2x
(iii) \(e^{\frac{x}{2}}\)
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 10
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 11

Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

Question 6.
Write the first 4 terms of the logarithmic series
(i) log(1 + 4x),
(ii) log(1 – 2x),
(iii) \(\log \left(\frac{1+3 x}{1-3 x}\right)\)
(iv) \(\log \left(\frac{1-2 x}{1+2 x}\right)\).
Find the intervals on which the expansions are valid.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 111
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 12
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 13

Question 7.
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 17
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 18

Question 8.
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 19
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 20

Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

Question 9.
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 21
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 22

Question 10.
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 23
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 24
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 25

Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 Additional Questions

Question 1.
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 26
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 27

Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

Question 2.
Write the four terms in the expansions of the following:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 28
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 29
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 30

Question 3.
Evaluate the following:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 31
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 32

Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

Question 4.
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 33
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 34

Question 5.
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 35
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 36

Samacheer Kalvi 11th Tamil Book Answers Solutions Guide

Expert Teachers at SamacheerKalvi.Guru has created Tamilnadu State Board Samacheer Kalvi 11th Tamil Book Answers and Solutions Guide Pdf Free Download are part of Samacheer Kalvi 11th Books Solutions. Here we have given TN State Board New Syllabus Samacheer Kalvi 11th Std Tamil Guide Pdf of Book Back Questions and Answers, Chapter Wise Important Questions, Study Material, Question Bank, Notes.

Students can also read Tamil Nadu Samacheer Kalvi 11th Tamil Model Question Papers 2020-2021.

Samacheer Kalvi 11th Tamil Guide Pdf Free Download

Tamilnadu State Board Samacheer Kalvi 11th Tamil Book Back Answers Solutions Guide.

Samacheer Kalvi 11th Tamil Book Back Answers

We hope the given Tamilnadu State Board Samacheer Kalvi 11th Tamil Book Solutions and Answers Pdf Free Download will help you. If you have any queries regarding TN State Board New Syllabus Samacheer Kalvi 11th Standard Tamil Guide Pdf of Book Back Questions and Answers, Chapter Wise Important Questions, Study Material, Question Bank, Notes, drop a comment below and we will get back to you at the earliest.

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5

Question 1.
Solve 2x2 + x – 15 ≤ 0.
Solution:
To find the solution of the inequality
ax2 + bx + c ≥ 0 or ax2 + bx +c ≤ 0 (for a > 0)
First we have to solve the quadratic equation ax2 + bx + c = 0
Let the roots be a and P (where a < P)
So for the inequality ax2 + bx + c ≥ 0 the roots lie outside α and β
(i.e.,) x ≤ α and x ≥ β
So for the inequality ax2 + bx + c ≤ 0. The roots lie between α and β
(i.e.,) x > α and x < β (i.e.) a ≤ x ≤ β
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5 1

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5

The inequality solver will then show you the steps to help you learn how to solve it on your own.

Question 2.
Solve -x2 + 3x – 2 ≥ 0
Solution:
-x2 + 3x – 2 ≥ 0 ⇒ x2 – 3x + 2 ≤ 0
(x – 1) (x – 2) ≤ 0
[(x – 1) (x – 2) = 0
⇒ x = 1 or 2.
Here α = 1 and β = 2. Note that α < β]
So for the inequality (x – 1) (x – 2) ≤ 2
x lies between 1 and 2
(i.e.) x ≥ 1 and x ≤ 2 or x ∈ [1, 2] or 1 ≤ x ≤ 2

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5 Additional Questions

Question 1.
Solve for x.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5 5
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5 6
Select the intervals in which (3x +1) (3x – 2) is positive
(3x + 1) > 0 and (3x – 2) > 0 or
3x +1 < 0 and 3x – 2 < 0
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5 7

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5

Question 2.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5 25
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5 26

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5

Question 3.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5 8
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5 9

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

Question 1.
Construct a quadratic equation with roots 7 and -3.
Solution:
Let the given roots be α = 7 and β = -3
Sum of the roots α + β = 7 + (-3)
α + β = 7 – 3 = 4
Product of the roots αβ = (7)(-3)
αβ = -21
The required quadratic equation is
x2 – (sum of two roots) x + Product of the roots = 0
x2 – 4x – 21 = 0

Question 2.
A quadratic polynomial has one of its zeros 1 + \(\sqrt{5}\) and it satisfies p(1) = 2. Find the quadratic polynomial.
Solution:
Given α = 1 + \(\sqrt{5}\) So, β = 1 – \(\sqrt{5}\)
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 1
The quadratic polynomial is
p(x) = x2 – (α + β)x + αβ
p(x) = k (x2 – 2x – 4)
p( 1) = k(1 – 2 – 4) = -5 k
Given p (1) = 2
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 2

The Polynomial Roots Calculator will find the roots of any polynomial with just one click.

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

Question 3.
If α and β are the roots of the quadratic equation x2 + \(\sqrt{2}\)x + 3 = 0, form a quadratic polynomial with zeroes 1/α, 1/β.
Solution:
α and β are the roots of the equation x2 + \(\sqrt{2}\)x + 3 = 0
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 3
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 4

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

Question 4.
If one root of k(x – 1)2 = 5x – 7 is double the other root, show that k = 2 or – 25.
Solution:
k(x – 1)2 = 5x – 7
(i.e.,) k(x2 – 2x + 1) – 5x + 7 = 0
x2 (k) + x(-2k – 5) + k + 1 = 0
kx2 – x(2k + 5) + (k + 7) = 0
Here it is given that one root is double the other.
So let the roots to α and 2α
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 5
2(4k2 + 25 + 20k) = 9k (k + 7)
2(4k2 + 25 + 20k) = 9k2 + 63k
8k2 + 50 + 40k – 9k2 – 63k = 0
-k2 – 23k + 50 = 0
k2 + 23k – 5o = 0
(k + 25)(k – 2) = 0
k = -25 or 2

Question 5.
If the difference of the roots of the equation 2x2 – (a + 1)x + a – 1 = 0 is equal to their product then prove that a = 2.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 6

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

Question 6.
Find the condition that one of the roots of ax2 + bx + c may be
(i) negative of the other
(ii) thrice the other
(iii) reciprocal of the other.
Solution:
(i) Let the roots be α and -β
Sum of the roots = – b/a = 0 ⇒ b = 0

(ii) Let the roots be α, 3α
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 7

Question 7.
If the equations x2 – ax + b = 0 and x2 – ex + f = 0 have one root in common and if the second equation has equal roots that ae = 2(b + f).
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 8

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

Question 8.
Discuss the nature of roots of
(i) -x2 + 3x + 1 = 0
(ii) 4x2 – x – 2 = 0
(iii) 9x2 + 5x = 0
Solution:
(i) -x2 + 3x + 1 = 0
x2 – 3x – 1 = 0 ———- (1)
Compare this equation with the equation
ax2 + bx + c = 0 ——– (2)
we have a = 1, b = -3, c = -1
Discriminant = b2 – 4ac
b2 – 4ac = (-3)2 – 4 × 1 × – 1
= 9 + 4 =13
b2 – 4ac = 13 > 0
∴ The two roots are real and distinct.

(ii) 4x2 – x – 2 = 0
4x2 – x – 2 = 0 ——(3)
Compare this equation with the equation
ax2 + bx + c = 0 (4)
we have a = 4 , b = – 1, c = – 2
Discriminant = b2 – 4ac
b2 – 4ac = (-1)2 – 4 (4) (-2)
= 1 + 32
= 33
b2 – 4ac = 33 >0
∴ The two roots are real and distinct.

(iii) 9x2 + 5x = 0
9x2 + 5x = 0 ——- (5)
Compare this equation with the equation
ax2 + bx + c = 0 ——– (6)
we have a = 9, b = 5 , c = 0
Discriminant = b2 – 4ac
b2 – 4ac = 52 – 4 × 9 × 0
b2 – 4ac = 25 > 0
∴ The two roots are real and distinct.

Question 9.
Without sketching the graphs find whether the graphs of the following functions will intersect the x- axis and if so in how many points.
(i) y = x2 + x + 2
(ii) y = x2 – 3x – 1
(iii) y = x2 + 6x + 9
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 20
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 21

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

Completing the Square Calculator is a free online tool that displays the variable value for the quadratic equation using completing the square method.

Question 10.
Write f(x) = x2 + 5x + 4 in completed square form.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 22

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 Additional Questions

Question 1.
Find the values of k so that the equation x2 = 2x (1 + 3k) + 7(3 + 2k) = 0 has real and equal roots.
Solution:
The equation is x2 – x(2) (1 + 3k) + 7 (3 + 2k) = 0
The roots are real and equal
⇒ ∆ = 0 (i.e.,) b2 – 4ac = 0
Here a = 1, b = -2 (1 + 3k), c = 7(3 + 2k)
So b2 – 4ac = 0
⇒ [-2 (1 + 3k)]2 – 4(1) (7) (3 + 2k) = 0
(i.e.,) 4 (1 + 3k)2 – 28 (3 + 2k) = 0
(÷ by 4) (1 + 3k)2 – 7(3 + 2k) = 0
1 + 9k2 + 6k – 21 – 14k = 0
9k2 – 8k – 20 = 0
(k – 2)(9k + 10) = 0
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 23
To solve the quadratic inequalities ax2 + bx + c < 0 (or) ax2 + bx + c > 0

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

Question 2.
If the sum and product of the roots of the quadratic equation ax2 – 5x + c = 0 are both equal to 10 then find the values of a and c.
Solution:
The given equation is ax2 – 5x + c = 0
Let the roots be α and β Given α + β = 10 and αβ = 10
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 24

Question 3.
If α and β are the roots of the equation 3x2 – 4x + 1 = 0, form the equation whose roots are \(\frac{\alpha^{2}}{\beta}\) and \(\frac{\beta^{2}}{\alpha}\)
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 25
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 26

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

Question 4.
If one root of the equation 3x2 + kx – 81 = 0 is the square of the other then find k.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 27

Question 5.
If one root of the equation 2x2 – ax + 64 = 0 is twice that of the other then find the value of a.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4 28

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9

Get the free “Partial Fraction Calculator” widget for your website, blog, WordPress, Blogger, or iGoogle.

Resolve the following rational expressions into partial fractions

Question 1.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 1
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 2

Question 2.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 3
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 4

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9

Question 3.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 5
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 45
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 55

Question 4.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 6
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 7

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9

Question 5.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 8
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 9
Equating nuemarator on bothsides we get
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 98

Question 6.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 146
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 11
Equating numerator on both sides
(x – 2)2 = A(x2 + 1) + (Bx + c)(x)
Put x = 0
1 = A
Equating co-eff of x2
1 = A + B
(i.e.,) 1 + B = 1 ⇒ B = 0
put x = 1
A(2) + B + C = 0 (i.e.,) 2A + B + C = 0
2 + 0 + C = 0 ⇒ C = -2
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 12

Question 7.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 13
Solution:
Since numerator and denominator are of same degree
we have divide the numerator by the denominator
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 14
Substituting the value in ….(1)
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 145

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9

Question 8.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 15
Solution:
Numerator is of greater degree than the denominator
So dividing Numerator by the denominator
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 16
⇒ 21x + 31 = A(x + 3) + B(x + 2)
Put x = -3
-63 + 31 = B(-1)
B = 32
Put x = -2
-42 + 31 = A(1) + B(0)
A = -11
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 17

Question 9.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 18
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 19
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 20

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9

Question 10.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 21
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 22
Equating Numerator on both sides we get
6x2 – x + 1 = A(x2 + 1) + (Bx + c)(x + 1)
6 + 1 + 1 = A(2) + 0 ⇒ 2A = 8 ⇒ A = 4
Equating co-eff of x2
6 = A + B
(i.e.,) 4 + B = 6 ⇒ B = 6 – 4 = 2
put x = 0
1 = A+ C
4 + C = 1 ⇒ C = 1 – 4 = -3
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 23

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9

Question 11.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 24
Solution:
Since Numerator and are of same degree divide Numerator by the denominator
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 25
equating Numerator on both sides we get
x – 5 = A(x + 3) + B(x – 1)
Put x = -3
-3 -5 = A(0) + B(-4)
-4B = -8 ⇒ B = 2
Put x = 1
1 – 5 = A(4) + B(0)
4A = -4 ⇒ A = -1
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 26

Question 12.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 27
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 28

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 Additional Questions

Question 1.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 29
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 30

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9

Question 2.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 31
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 32

Question 3.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 33
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 34
Equating nemerator on b/s
9 = A(x+2)2 + B(x – 1)(x + 2) + C(x – 1)
Put x = -2
9 = A(0) + B(0) + C(-3)
-3C = 9 ⇒ C = -3
Put x = 1
9 = A (1 + 2)2 + B (0) + C(0)
9A = 9
A = 1
Put x = 0
9 = 4A – 2B – C
9 = 4(1) – 2B + 3
9 – 7 = -2B
2 = -2B
B = -1
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 35

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9

Question 4.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 36
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 133

Question 5.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 38
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 39
0 = 0 + B(1 + 2)
3B = 0 ⇒ B = 0
Put x = -2
(-2)3 – 1 = A(-2 – 1) + B(0)
-8 – 1 = -3A
-9 = -3A
A = 9/3 ⇒ A = 3
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 40

Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy

Students can Download Economics Chapter 7 Indian Economy Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Economics Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy

Samacheer Kalvi 11th Economics Indian Economy Text Book Back Questions and Answers

MCQ Questions for Class 11 Hindi with Answers – NCERT Solutions.

Part – A

Multiple Choice Questions

Question 1.
The main gold mine region in Karnataka is _________
(a) Kolar
(b) Ramgiri
(c) Anantpur
(d) Cochin
Answer:
(a) Kolar

Question 2.
The economic growth of a country is measured by national income indicated by _________
(a) GNP
(b) GDP
(c) NNP
(d) Per capita income
Answer:
(b) GDP

Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy

Question 3.
Which one of the following are developed nations?
(a) Mexico
(b) Ghana
(c) France
(d) SriLanka
Answer:
(c) France

Question 4.
The position of the Indian Economy among the other strongest economies in the world is _______
(a) Fourth
(b) Seventh
(c) Fifth
(d) Tenth
Answer:
(b) Seventh

Question 5.
Mixed economy means _________
(a) Private sectors and banks
(b) Co-existence of Public and Private sectors
(c) Public sectors and banks
(d) Public sectors only
Answer:
(b) Co-existence of Public and Private sectors

Question 6.
The weakness of Indian Economy is _________
(a) Economic disparities
(b) Mixed economy
(c) Urbanisation
(d) Adequate employment opportunities
Answer:
(a) Economic disparities

Question 7.
A scientific study of the characteristics of population is _________
(a) Topography
(b) Demography
(c) Geography
(d) Philosophy
Answer:
(b) Demography

Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy

Question 8.
The year 1961 is known as _________
(a) Year of the small divide
(b) Year of Population Explosion
(c) Year of Urbanisation
(d) Year of Great Divide
Answer:
(b) Year of Population Explosion

Question 9.
In which year the population of India crossed the one billion mark?
(a) 2000
(b) 2001
(c) 2005
(d) 1991
Answer:
(b) 2001

Question 10.
The number of deaths per thousand population is called as ________
(a) Crude Death Rate
(b) Crude Birth Rate
(c) Crude Infant Rate
(d) Maternal Mortality Rate
Answer:
(a) Crude Death Rate

Question 11.
The number of births per thousand population is called as _________
(a) Crude death rate
(b) Mortality rate
(c) Morbidity rate
(d) Crude Birth Rate
Answer:
(d) Crude Birth Rate

Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy

Question 12.
Density of population = _________
(a) Land area / Total Population
(b) Land area / Employment
(c) Total Population / Land area of the region
(d) Total Population
Answer:
(c) Total Population / Land area of the region

Question 13.
Who introduced the National Development Council in India?
(a) Ambedkar
(b) Jawaharlal Nehru
(c) Radhakrishanan
(d) V.K.R.V. Rao
Answer:
(b) Jawaharlal Nehru

Question 14.
Who among the following propagated Gandhian Economic thinkings.
(a) Jawaharlal Nehru
(b) VKRV Rao
(c) JC Kumarappa
(d) A.K.Sen
Answer:
(c) JC Kumarappa

Question 15.
The advocate of democratic socialism was
(a) Jawaharlal Nehru
(b) P.C. Mahalanobis
(c) Dr. Rajendra Prasad
(d) Indira Gandhi
Answer:
(a) Jawaharlal Nehru

Question 16.
Ambedkar the problem studied by in the context of Indian Economy is ________
(a) Small land holdings and their remedies
(b) Problem of Indian Currency
(c) Economics of socialism
(d) All of them
Answer:
(b) Problem of Indian Currency

Question 17.
Gandhian Economics is based on the Principle _________
(a) Socialistic idea
(b) Ethical foundation
(c) Gopala Krishna Gokhale
(d) Dadabhai Naoroji
Answer:
(b) Ethical foundation

Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy

Question 18.
V.K.R.V Rao was a student of _________
(a) J.M. Keynes
(b) Colin Clark
(c) Adam smith
(d) Alfred Marshal
Answer:
(a) J.M. Keynes

Question 19.
Amartya Kumara Sen received the Nobel prize in Economics in the year.
(a) 1998
(b) 2000
(c) 2008
(d) 2010
Answer:
(a) 1998

Question 20.
Thiruvalluvar economic ideas mainly dealt with _________
(a) Wealth
(b) Poverty is the curse in the society
(c) Agriculture
(d) All of them
Answer:
(d) All of them

Part – B
Answer the following questions in one or two sentences

Question 21.
Write the meaning of Economic Growth.
Answer:

  1. Economic growth is usually measured by National Income, indicated by Gross Domestic Product. [GDP],
  2. The GDP is the total monetary value of the goods and services produced by that country over a specific period of time, usually one year.
  3. On the basis of the level of economic development, nations are classified as developed and developing economies.

Question 22.
State any two features of a developed economy.
Answer:

  1. High National Income
  2. High per capita Income

Question 23.
Write the short note on natural resources
Answer:

  1. Any stock or reserve that can be drawn from nature is a Natural Resources.
  2. The major natural resources are – land, forest, water, mineral, and energy.
  3. India is rich in natural resources, but the majority of the Indians are poor.
  4. Nature has provided with diverse climate, several rivers for irrigation and power generation, rich minerals, rich forest, and diverse soil.

Question 24.
Point out anyone feature of Indian Economy
Answer:
Indian economy is a mixed economy. In India, both the private and public sectors coexist.

Question 25.
Give the meaning of non-renewable energy
Answer:
Non-renewable energy sources

  1. The sources of energy which cannot be renewed or re-used are called non-renewable energy sources.
  2. Basically, these are the energy sources that will get exhausted over a period of time.
  3. Some of the examples of this kind of resources are coal, oil, gas, etc.

Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy

Question 26.
Give a short note on Sen’s ‘Choice of Technique’.
Answer:
In a labour surplus economy, the generation of employment cannot be increased at the initial stage by the adaptation of capital intensive technique what’s called the choice of technique.

Question 27.
List out the reasons for low per capita income as given by V.K.R.V. Rao.
Answer:

  1. Uneconomic holdings with sub-divisions and fragmentation.
  2. Low levels of water availability for crops.
  3. Excess population pressure on agriculture due to the absence of a large industrial sector.
  4. Absence of capital.
  5. Absence of autonomy in currency policy, and in general in monetary matters encouraging holding of gold.

Part – C
Answer the following questions in One Paragraph

Question 28.
Define Economic Development.
Answer:
The level of economic development is indicated not just by GDP, but by an increase in citizen’s quality of life or well-being. The quality of life is being assessed by several indices such as the Human Development Index (HDI), Physical Quality of Life Index (PQLI), and Gross National Happiness Index (GNHI)

Question 29.
State Ambedkar’s Economic ideas on agricultural economics.
Answer:
In thel918, Ambedkar published a paper “Small Holding in India and their Remedies”. Citing Adam Smith’s “Wealth of Nations”, he made a fine distinction between “Consolidation of Holdings” and Enlargement of Holdings”.

Question 30.
Write on the short note on village Sarvodaya.
Answer:
Village Sarvodaya is the concept of Gandhiji on the development of villages. To Gandhi, India lives in villages. He was interested in developing the villages as self-sufficient units.

According to him, “Real India was to be found in villages and not in towns or cities”.
So he suggested the development of self-sufficient, self-dependent villages.

Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy

Question 31.
Write the strategy of Jawahar lal Nehru in India’s planning.
Answer:

  1. Jawaharlal Nehru was responsible for the introduction of planning in our country.
  2. To Jawaharlal Nehru, the Plan was essentially an integrated approach for development.
  3. Initiating the debate on the Second Plan in the Lok Sabha in May 1956, Nehru spoke on the theme of planning.
  4. Nehru Said, “the essence of planning is to find the best way to utilize all resources of manpower, of money and so on”.
  5. Planning for Nehru was essentially linked up with industrialization and eventual self-reliance for the country’s economic growth on a self-accelerating growth.
  6. Nehru carried through this basic strategy of planned development.

Question 32.
Write the V.K.R.V.Rao’s contribution to the multiplier concept.
Answer:

  1. Rao’s examination of the “interrelation between investment, income and multiplier in an underdeveloped economy (1952)” was his major contribution to macroeconomic theory.
  2. In it, he asserts that the Keynesian multiplier principle remains inoperative in underdeveloped countries.
  3. V.K.R.V. Rao was the best equipped of all Keynes’ pupils.

Question 33.
Write a short note on Welfare Economics given by Amartya Sen.
Answer:

  1. Amartya Kumar Sen has included the concept of entitlement items like nutrition, food, medical and health care, employment, the security of food supply in times of famine, etc.
  2. He considered famine as arising out of the failure of establishing a system of entitlements.

Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy

Question 34.
Explain Social infrastructure.
Answer:

  1. Social infrastructure refers to those structures which are improving the quality of man power and contribute indirectly towards the growth of an economy.
  2. These structures are outside the system of production and distribution.
  3. The development of these social structures help in increasing the efficiency and productivity of man power.
    (Eg.) Schools, Colleges, Hospitals

Part – D
Answer the following questions in about a page

Question 35.
Explain strong features Indian economy
Answer:
Strengths of Indian Economy:
1. India has a mixed economy:

  • Indian economy is a typical example of mixed economy.
  • This means both private and public sectors co-exist and function smoothly.
  • The fundamental and heavy industrial units are being operated under the public sector.
  • Due to the liberalization of the economy, the private sector has gained importance.
  • This makes it a perfect model for public-private partnership.

2. Agriculture plays the key role:

  • Agriculture being the maximum pursued occupation in India.
  • It plays an important role in its economy as well.
  • Around 60% of the people in India depend upon agriculture for their livelihood.
  • In fact, about 17% of our GDP today is contributed by the agriculture sector.
  • Green revolution, ever green revolution and inventions is bio technology have made agriculture self sufficient and also surplus production.
  • The export of agricultural products such as fruits, vegetables, spices, vegetable oils, tobacco, animal skin, etc. also add to forex earning through international trading.

3. An emerging market:

  • Indian has emerged as vibrant economy sustaining stable GDP growth rate even in the midst of global downtrend.
  • This has attracted significant foreign capital through FDI and FII.
  • India has a high potential for prospective growth.
  • This also makes it an emerging market for the world.

4. Emerging Economy:

  • Emerging as a top economic giant among the world economy.
  • India bags the seventh position in terms of nominal Gross Domestic Product [GDP] and third in terms of Purchasing Power Parity [PPP].
  • As a result of rapid economic growth, Indian economy has a place among the G20 countries.

5. Fat Growing Economy:

  • Indians economy is well known for high and sustained growth.
  • It has emerged as the world’s fastest growing economy in the year 2016-17 with the growth rate of 7.1 % in GDP.

6. Fast Growing Service Sector:

  • The service sector, contributes a lion’s share of the GDP in India.
  • There has been a high rise growth in the technical sectors like information technology.

7. Large Domestic Consumption:

  • With the faster growth rate in the economy the standard of living has improved a lot.
  • The standard of living has considerably improved and life style has changed.

8. Rapid Growth of Urban Areas:

  • Urbanization is a key ingredient of the growth of any economy.
  • Improved connectivity is transport and communication, education and health have speeded up the pace of urbanization.

9. Stable Macro Economy:

  • The Indian economy has been projected and considered as one of the most stable economies of the world.
  • The current year’s Economic survey represents the Indian economy to be a “heaven of macroeconomic stability, resilience and optimism”.

10. Demographic dividend:

  • This means that India is a pride owner of the maximum percentage of youth.
  • The young population is not only motivated but skilled and trained enough to maximize ‘ the growth.
  • Thus human capital plays a key role in maximizing the growth prospects is the country.
  • This has invited foreign investments to the country and outsourcing opportunities too.

Question 36.
Write the importance of mineral resources in India.
Answer:

  1. Iron-ore : Hematite iron is mainly found in Chattisgarh, Jharkhand, Odisha, Goa and Karnataka. Magnetite is found in Western Coast of Karnataka. Some deposits of iron ore are also found in Kerala, Tamil Nadu and Andhra Pradesh.
  2. Coal and Lignite : India ranks third in the coal production. The main centres are West-Bengal, Bihar, Madhya Pradesh, Maharashtra, Odisha and Andhra Pradesh. Bulk production comes from Bengal-Jharkhand coal fields. Lignite is found in Neyveli in Tamil Nadu.
  3. Bauxite : Bauxite is a main source of aluminium. Major reserves are found in Odisha and Andhra Pradesh.
  4. Mica : India stand first in sheet mica production. Andhra Pradesh, Jharkhand, Bihar and Rajasthan are major reserves.
  5. Crude Oil : Oil is being explored in India at many places of Assam and Gujarat.
  6. Gold : India possesses only a limited gold reserve.
  7. Diamond: The total reserves of a diamond is estimated at around 4582 carats.

Question 37.
Bring out Jawaharlal Nehru’s contribution to the idea of economic development.
Answer:

  1. Jawaharlal Nehru, one of the chief builders of Modem India, was the first Prime Minister of Independent India.
  2. He was a great patriot, thinker, and statesman.

1. Democracy and Secularism:

  • Jawaharlal Nehru was a firm believer in democracy.
  • He believed in free speech civil liberty, adult franchise and the Rule of Law and Parliamentary democracy.
  • Secularism is another signal contribution of Nehru to india.

2. Planning:

  • Jawaharlal Nehru was responsible for the introduction of planning in our country.
  • Jawaharlal Nehru, the Plan was essentially an integrated approach for development.
  • Nehru Spoke on the theme of planning.
  • He said “the essence of planning is to find the best way to utilize all resources of manpower, of money and so on”.
  • “Planning for Nehru was essentially linked up with Industrialization and eventual self-reliance for the country’s economic growth on a self-accelerating growth.
  • Nehru carried through this basis strategy of planned development.
  • Nehru’s contribution to the advancement of science, research, technology and industrial development cannot be forgotten.
  • It was during is period, many IITs and Research Institutions were established.
  • Nehru always in insited on “ scientific temper”.

Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy

Question 38.
Write a brief note on the Gandhian economic ideas.
Answer:
Gandhian economics is based on ethical foundations.
Salient features of Gandhian economic thought:

  1. Village republics : Gandhi was interested in developing the villages as self-sufficient units.
  2. On machinery : Gandhi described machinery as ‘Great Sin’.
  3. Industrialism : Gandhi considered industrialism as a curse on mankind.
  4. Decentralization : He advocated a decentralized economy.
  5. Village sarvodaya : Gandhi suggested the development of self-sufficient, self-dependent villages.
  6. Bread labour: Gandhi realized the dignity of human labour. Bread labour or body labour was the expression that Gandhi used to mean manual labour.
  7. The doctrine of trusteeship : Trusteeship provides a means of transforming the present capitalist society into an egalitarian one.
  8. On the food problem : Gandhi was against any sort of food controls. Food controls only create artificial scarcity.
  9. On population : Gandhi was in favour of birth control through brahmacharya or self-control.
  10. On prohibition: Gandhi advocated cent percent prohibition.

Samacheer Kalvi 11th Economics Indian Economy Additional Questions and Answers

Part -A
Choose the best options

Question 1.
Third world country __________
(a) India
(b) China
(c) Bangladesh
(d) Srilanka
Answer:
(c) Bangladesh

Question 2.
__________ is the year of great divide.
(a) 1921
(b) 1918
(c) 1943
(d) 1969
Answer:
(a) 1921

Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy

Question 3.
__________ are BIMARU States.
(a) Tamil Nadu, Kerala, Karnataka, Madhya Pradesh
(b) Bihar, Madhya Pradesh, Rajasthan, Uttarpradesh
(c) West Bengal, Himachal Pradesh, Punjab, Delhi
(d) Andhra Pradesh, Odisa, Gujarat, Chattisgarh
Answer:
(b) Bihar, Madhya Pradesh, Rajasthan, Uttarpradesh

Question 4.
In India, Lignite is found in.
(a) New Delhi
(b) Kolkatta
(c) Ranji
(d) Neyveli
Answer:
(d) Neyveli

Question 5.
Indian educational system comprises of __________ stages.
(a) 6
(b) 12
(c) 3
(d) 4
Answer:
(a) 6

Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy

Question 6.
__________ is considered as the best pupil of J.M. Keynes
(a) A.K. Sen
(b) J.C.Kumarappa
(c) V.K.R.V. Rao
(d) Manmohan Singh
Answer:
(c) V.K.R.V. Rao

Question 7.
Indian Railways introduced first wifi facility in __________
(a) Chennai
(b) Calcutta
(c) Bengaluru
(d) Hyderabad
Answer:
(c) Bengaluru

Question 8.
National Harbour board was set up in __________
(a) 1947
(b) 1950
(c) 1948
(d) 1951
Answer:
(b) 1950

Question 9.
A study on poverty and famines and the concept of entitlements and capability development was done by __________
(a) Adamsmith
(b) A.K.Sen
(c) Marshall
(d) V.K.R.V. Rao
Answer:
(b) A.K.Sen

Question 10.
Percapital income is __________
(a) National income / output
(b) National income / population
(c) National income / percapital output
(d) National income / male population
Answer:
(b) National income / population

Question 11.
‘India lives in villages’ said by __________
(a) V.K.R.V. Rao
(b) J.C. Kumarappa
(c) Mahatma Gandhi
(d) Jawaharlal Nehru
Answer:
(c) Mahatma Gandhi

Match the following and choose the answer using the codes given below

Question 1.
Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy
(a) 1 2 3 4
(b) 3 4 2 1
(c) 1 3 4 2
(d) 2 4 3 1
Answer:
(b) 3 4 2 1

Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy

Question 2.
Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy
(a) 4 1 2 3
(b) 3 4 2 1
(c) 2 4 3 1
(d) 1 2 4 3
Answer:
(a) 4 1 2 3

Question 3.
Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy
(a) 4 1 2 3
(b) 1 2 3 4
(c) 4 3 2 1
(d) 2 4 1 3
Answer:
(d) 2 4 1 3

Choose the odd one out

Question 4.
(a) Poverty and famine
(b) Choice of technique
(c) Economics of caste
(d) Concept of capability
Answer:
(c) Economics of caste

Question 5.
(a) Budget by assignment
(b) Budget by assigned revenue
(c) Budget by shared revenues
(d) Budget by fiscal revenues
Answer:
(d) Budget by fiscal revenues

Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy

Question 6.
(a) Net Domestic Product (NDP)
(b) Human Development Index (HDI)
(c) Physical Quality of Life Index (PQLI)
(d) Gross National Happiness Index (GNHI)
Answer:
(a) Net Domestic Product (NDP)

Choose the correct statement

Question 7.
(a) Indian economy is the seventh-largest in the world
(b) Indian economy is a mixed economy
(c) India ranks third in terms of purchasing power parity
(d) The human capital of India is old
Answer:
(d) The human capital of India is old

Question 8.
(a) The annual addition of population in India equals the total population of Australia.
(b) There exists a huge economic disparity in the Indian economy
(c) Kerala has the highest birth rate
(d) Every 6th person in the world is an Indian
Answer:
(c) Kerala has the highest birth rate

Choose the incorrect statement

Question 9.
Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy
Answer:
(b) Demographic transition – (ii) 2001

Question 10.
Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy
Answer:
(c) 1971 – (iii) 1040

Choose the correct statement

Question 11.
(a) Nehru always insisted on scientific temper
(b) Gandhi advocated a centralized economy
(c) Valluvar has recommended an unbalanced budget
(d) V.K.R.V. Rao’s economic thought is coined as Gandhian Economics.
Answer:
(a) Nehru always insisted on scientific temper

Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy

Question 12.
(a) Until 1979, Education in India was in state list
(b) Education in India follows the 10+3 pattern
(c) Sex ratio refers to the number of females per 1000 males.
(d) India ranks fourth in coal production of the world.
Answer:
(c) Sex ratio refers to the number of females per 1000 males.

Analyze the reason for the following 

Question 13.
Assertion (A) : There exists a huge economic disparity in the Indian economy.
Reason (R) : The proportion of income and assets owned by top 10% of Indians goes on increasing.
(a) Both (A) and (R) are true, (R) is the correct explanation of (A)
(b) Both (A) and (R) are true, (R) is the not correct explanation of (A)
(c) (A) is true and (R) are false.
(d) (A) is false, but (R) is true.
Answer:
(a) Both (A) and (R) are true, (R) is the correct explanation of (A)

Question 14.
Assertion (A) : In Kerala, the adult sex ratio is 1084 as on 2011.
Reason (R) : Kerala provides better status to women as compared to other states.
(a) Both (A) and (R) are true, (R) is not the correct explanation of (A)
(b) Both (A) and (R) are true, (R) is the correct explanation of (A)
(c) Both (A) and (R) are false.
(d) None of the above
Answer:
(b) Both (A) and (R) are true, (R) is the correct explanation of (A)

Fill in the blanks with suitable option given below

Question 15.
In India, Lignite is found in _____
(a) New Delhi
(b) Kolkata
(c) Ranji
(d) Neyveli
Answer:
(d) Neyveli

Question 16.
Indian educational system comprises of _____ stages
(a) 12
(b) 3
(c) 6
(d) 4
Answer:
(c) 6

Question 17.
Indian Railways introduced first wifi facility in _____
(a) Chennai
(b) Bengaluru
(c) Calcutta
(d) Hyderabad
Answer:
(b) Bengaluru

Choose the best option 

Question 18.
_____ is considered as the best pupil of J.M. Keynes
(a) A.K. Sen
(b) J.C.Kumarappa
(c) V.K.R.V.Rao
(d) Nehru
Answer:
(c) V.K.R.V.Rao

Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy

Question 19.
Third world country is _____
(a) India
(b) China
(c) Srilanka
(d) Bangladesh
Answer:
(d) Bangladesh

Question 20.
Per capita income is _____
(a) National income / output
(b) National income / population
(c) National income / per capital output
(d) National income / male population
Answer:
(b) National income / population

Part – B
Answer the following questions in one or two sentences 

Question 1.
What is Gross National Happiness Index?
Answer:
GNHI is an indicator of progress, which measures sustainable development, environmental conservation, promotion of culture and good governance.

Question 2.
State any two weaknesses of Indian Economy.
Answer:

  1. Large population.
  2. Inequality and poverty.

Question 3.
What is sex ratio?
Answer:
Sex ratio refers to the number of females per 1,000 males.

Question 4.
Explain economic infrastructure.
Answer:
Economic infrastructure is the support system which helps in facilitating the production

Question 5.
What is Gandhian Economics?
Answer:
J.C. Kumarappa developed economic theories based on Gandhism. He developed a school of economic thought coined as Gandhian Economics.

Question 6.
Which are “BIMARU” states ?
Answer:

  1. Bihar
  2. Madhya Pradesh
  3. Rajasthan
  4. Uttarpradesh

Question 7.
What is year of small divide?
Answer:
In 1951, population growth rate has come down from 1.33% to 1.25%. Hence it is known as year of small divide.

Part – C
Answer the following questions in one Paragraph

Question 1.
What are the aspects of demographic trends in India ?
Answer:

  1. Size of population
  2. Rate of growth.
  3. Birth and death rates
  4. Density of population
  5. Sex – ratio
  6. Life – expectancy at birth
  7. Literacy ratio.

Question 2.
Explain Thiruvalluvar’s ideas on agriculture.
Answer:
According to Thiruvalluvar, agriculture is the most fundamental economic activity. They are the axle-pin of the world, for on their prosperity revolves prosperity of other sectors of the economy. He says, plowmen alone is the freemen of the soil. He believes that agriculture is superior to all other occupations.

Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy

Question 3.
Write short note on Birth rate and Death rate.
Answer:
Short Note On Birth Rate And Death Rate:

  1. Birth Rate : It refers to the number of births per thousand of the population.
  2. Death Rate : It refers to the number of deaths per thousand of the population.

Kerala has the lowest birth rate and Uttarpradesh has the highest birth rate. West Bengal has the lowest and Orissa has the highest death rates.

Part – D
Answer the following questions in about a page

Question 1.
Explain about health in India.
Answer:
Health in India is a state government responsibility. The central council of health and welfare formulates the various health care projects and health department reform policies. The administration of health industry is the responsibility of the ministry of health and welfare.

Health care in India has many forms. But, all medical systems are under one ministry AYUSH.
Health status is better in Kerala as compared to other states. India’s health status is poor compared to Sri Lanka.

Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy

Question 2.
Explain Thiruvalluvar’s economic ideas.
Answer:
The economic ideas of Thiruvalluvar are found in his immortal work, Thirukkural in its second part porutpal.

  1. Factors of production: Thiruvalluvar has made many passing references about the factors of production, land, labour, capital, organization.
  2. Agriculture: Agriculture is the most fundamental economic activity. The prosperity of other sectors depends on the prosperity of agriculture. Valluvar believes that agriculture is superior to all other occupations.
  3. Public finance: He elaborately explained public finance under the headings public revenue, financial administration and public expenditure.
  4. Public expenditure: Valluvar has recommended a balanced budget. He advocates
    • Defence
    •  Public works
    • Social Services
  5. External assistance: Valluvar was against seeking external assistance. He advocated a self-sufficient economy.
  6. Poverty and begging: According to him ‘poverty is the root cause of all other evils which would lead to ever-lasting sufferings.
  7. Wealth: Valluvar has regarded wealth as only a means and not an end.
  8. Welfare state: The important elements of a welfare state are
    • Perfect health of the people without the disease
    • Abundant wealth
    • Good crop
    • Prosperity and happiness
    • Full security for the people.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Students can Download Chemistry Chapter 9 Solutions Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Chemistry Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Samacheer Kalvi 11th Chemistry Chapter 9 Solutions Textual Evaluation Solved

Samacheer Kalvi 11th Chemistry Solutions Multiple Choice Questions

Question 1.
The molality of a solution containing 1 .8g of glucose dissolved in 250g of water is …………
(a) 0.2 M
(b) 0.01 M
(c) 0.02 M
(d) 0.04 M
Answer:
(d) 0.04 M
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-1

Question 2.
Which of the following concentration terms is/are independent of temperature?
(a) molality
(b) molarity
(c) mole fraction
(d) (a) and (c)
Answer:
(d) (a) and (c)
Solution:
Molality and mole fraction are independent of temperature.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 3.
Stomach acid, a dilute solution of HCI can be neutralised by reaction with Aluminium hydroxide
Al(OH)3 + 3HCl(aq) → AlCl3 + 3H2O
How many millilitres of 0.1 M Al(OH)3 solution are needed to neutralise 21 mL of 0.1 M HCl
(a) 14 mL
(b) 7 mL
(c) 21 mL
(d) none of these
Answer:
(b) 7 mL
Solution:
M1 x V1 = M2 x V2
∵ 0.1 M Al(OH)3 gives 3 x 0.1 = 0.3 M OH ions .
0.3 x V1 = 0.1 x 21
V1= \(\frac { 0.1 x 21 }{ 0.3 }\) = 7ml

Question 4.
The partial pressure of nitrogen in air is 0.76 atm and its Henry’s law constant is 7.6 x 104 atm at 300K. What is the mole fraction of nitrogen gas in the solution obtained when air is bubbled through water at 300K?
(a) 1 x 10-4
(b) 1 x 10-6
(c) 2 x 10-5
(d) 1 x 10-5
Answer:
(d) 1 x 10-5
Solution:
PN2 = 0.76atm
KH = 7.6 x 104
x = ?
PN2 = KH . x
0.76 = 7.6 x 104x x
x = \(\frac { 0.76 }{ 7.6\times { 10 }^{ 4 } }\) = 1 x 10-5

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 5.
The Henry’s law constant for the solubility of Nitrogen gas in water at 350K is 8 x 104 atm. The mole fraction of nitrogen in air is 0.5. The number of moles of Nitrogen from air dissolved in 10 moles of water at 350K and 4 atm pressure is ………….
(a) 4 x 10-4
(b) 4 x 104
(c) 2 x 10-2
(d) 2.5 x 10-4
Answer:
(d) 2.5 x 10-4
Solution:
KH = 8 x 104
(xN2 )in air = 0.5
Total pressure = 4 atm
Partial pressure of nitrogen = Mole fraction Total pressure
= O.5 x 4 = 2
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-3

Question 6.
Which one of the following is incorrect for ideal solution?
(a) ∆Hmix = 0
(b) ∆Umix = o
(c) ∆P = PObserved – PCalculated by raoults law = 0
(d) ∆Gmix = 0
Answer:
(d) ∆Gmix = 0
Solution:
For an ideal solution, ∆Smix \(\neq\) 0; Hence ∆Gmix \(\neq\) 0
∴ Incorrect is ∆Gmix = 0

Question 7.
Which one of the following gases has the lowest value of Henry’s law constant?
(a) N2
(b) He
(c) CO2
(d) H2
Answer:
(c) CO2
Solution:
Carbon dioxide; most stable gas and has lowest value of Henry’s Law constant.

Question 8.
P1 and P2 are the vapour pressures of pure liquid components, 1 and 2 respectively of an ideal binary solution if x1 represents the mole fraction of component 1, the total pressure of the solution formed by 1 and 2 will be ………
(a) P1 + x1(P2 – P1)
(b) P2 – x1(P2 + P1)
(c) P1 – x2(P1 – P2)
(d) P1 + x2(P1 – P2)
Answer:
(c) P1 – x2(P1 – P2)
Solution:
Ptotal = P1 + P2
= P1 x1 + P2x2
= P1(1 – x2) + P2x2
= P1 – P1x2 + P2x2 = P1 – x2(P1 – P2)
[∵x1 + x2 = 1
x1 = 1 – x2]

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 9.
Osomotic pressure (π) of a solution is given by the relation ……………
(a) π = nRT
(b) πV = nRT
(c) πRT = n
(d) none of these
Answer:
(b) πV = nRT
Solution:
n = CRT
n = \(\frac { n }{ V }\)
π V = nRT

Question 10.
Which one of the following binary liquid mixtures exhibits positive deviation from Raoults law?
(a) Acetone + chloroform
(b) Water + nitric acid
(c) HCI + water
(d) ethanol + water
Answer:
(d) ethanol + water

Question 11.
The Henry’s law constants for two gases A and B are x and y respectively. The ratio of mole fractions of A to B is 0.2. The ratio of mole fraction of B and A dissolved in water will be …………
(a) \(\frac { 2x }{ y }\)
(b) \(\frac { y }{ 0.2x }\)
(c) \(\frac { 0.2x }{ y }\)
(d) \(\frac { 5x }{ y }\)
Answer:
(d) \(\frac { 5x }{ y }\)
Solution:
Given,
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-4

Question 12.
At 100°C the vapour pressure of a solution containing 6.5g a solute in 100g water is 732mm. If Kb = 0.52, the boiling point of this solution will be …………..
(a) 102°C
(b) 100°C
(c) 101°C
(d) 100.52°C
Answer:
(c) 101°C
Solution:
\(\frac { ΔP }{ P° }\) = \(\frac { { n }_{ 2 } }{ { n }_{ 1 } }\)
W2 = 6.5g
W1 = 100g
Kb = 0.52
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-5
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-6

Question 13.
According to Raoult’s law, the relative Lowering of vapour pressure for a solution is equal to…
(a) mole fraction of solvent
(b) mole fraction of solute
(c) number of moles of solute
(d) number of moles of solvent
Answer:
(b) mole fraction of solute
Solution:
\(\frac { ∆P }{ P° }\) = x2 (Mole fraction of the solute)

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 14.
At same temperature. which pair of the following solutions are isotonic?
(a) 0.2 M BaCl2 and 0.2M urea
(b) 0.1 M glucose and 0.2 M urea
(c) 0.1 MNaCl and 0.1 MK2SO4
(d) 0.1 MBa(NO3)2 and 0.1 MNa2 SO4
Answer:
(d) 0.1 M Ba (NO3)2 and 0.1 M Na2 SO4
Solution:
0.1 x 3 ion [Ba2 + 2NO3], 0.1 x 3 ion [2Na+, SO4]

The formula of normality a solution is the gram equivalent weight of a solute per liter of solution.

Question 15.
The empirical formula of a non-electrolyte(X) is CH2O. A solution containing six gram of X exerts the same osmotic pressure as that of 0.025 M glucose solution at the same temperature. The molecular formula of X is
(a) C2H4O2
(b) C8H16O8
(c) C4H8O4
(d) CH2O
Answer:
(b) C8H16O8
Solution:
1)non electrolute = (π2)glucose
C1RT = C2RT
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-7
\(\frac { 6 }{ n(30) }\) = 0.025
n = \(\frac { 6 }{ 0.025 x 30 }\) = 30
∴ Molecular formula C8H16O8

Question 16.
The KH for the solution of oxygen dissolved in water is 4 x 104 atm at a given temperature. If the partial pressure of oxygen in air is 0.4 atm, the mole fraction of oxygen in solution is …………..
(a) 4.6 x 103
(b) 1.6 x 104
(c) 1 x 10-5
(d) 1 x 105
Answer:
(c) 1 x 10-5
Solution:
KH = 4 x 104 atm,
(PO2)air = 0.4 atm,
(xo2)in solution = ?
air – in solution
(PO2)air = KH(xo2)in solution
0.4 = 4 x 104(xo2)in solution
(xo2)in solution = \(\frac { 0.4 }{ 4\times { 10 }^{ 4 } }\) = 1 x 10-5

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 17.
Normality of 1.25M sulphuric acid is …………
(a) 1.25 N
(b) 3.75 N
(c) 2.5 N
(d) 2.25 N
Answer:
(c) 2.5 N
Solution:
Normality of H2SO4 = (No. of replacable H+) x M = 2 x 1.25 = 2.5 N

Question 18.
Two liquids X and Y on mixing gives a warm solution. The solution is …………..
(a) ideal
(b) non-ideal and shows positive deviation from Raoults law
(c) ideal and shows negative deviation from Raoults Law
(d) non – ideal and shows negative deviation from Raoults Law
Answer:
(d) non – ideal and shows negative deviation from Raoults Law
Solution:
∆Hmix is negative and show negative deviation from Raoults law.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 19.
The relative lowering of vapour pressure of a sugar solution in water is 3.5 x 10-3. The mole fraction of water in that solution is …………
(a) 0.0035
(b) 0.35
(c) 0.0035/18
(d) 0.9965
Answer:
(d) 0.9965
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-9

Question 20.
The mass of a non-volatile solute (molar mass 80 g mol-1) which should be dissolved in 92g of toluene to reduce its vapour pressure to 90% ………..
(a) 10g
(b) 20g
(c) 9.2 g
(d) 8.89g
Answer:
(d) 8.89g
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-10
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-11

Question 21.
For a solution, the plot of osmotic pressure (π) verses the concentration (e in mol L-1) gives a straight line with slope 310 R where ‘R’ is the gas constant. The temperature at which osmotic pressure measured is ………..
(a) 310 x 0.082 K
(b) 3 10°C
(c) 37°C
(d) \(\frac { 310 }{ 20.082}\)
Answer:
(c) 37°C
Solution:
π = CRT
y = x(m)
m = RT
310 R = RT
T = 310 K
= 37°C

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 22.
200 ml of an aqueous solution of a protein contains 1 .26g of protein. At 300K, the osmotic pressure of this solution is found to be 2.52 x 10-3 bar. The molar mass of protein will be (R =0.083 Lhar mol-1 K-1) ……………
(a) 62.22 Kg mol-1
(b) 12444 g mol-1
(c) 300g mol-1
(d) none of these
Answer:
(a) 62.22 Kg mol-1
Solution:
π = CRT
M = \(\frac { WRT }{ π1 }\) = \(\frac { 1.26\times 0.083\times 300 }{ 2.52\times { 10 }^{ -3 }\times 0.2 }\) = 62.22Kg mol-1

Question 23.
The Van’t Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is ………..
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(b) 1
Solution:
Ba(OH)2 dissociates to form Ba2+ and 2OH-1 ion
α = \(\frac { (i – 1) }{ (n – 1) }\)
i = α (n – 1) + 1
n = i = 3 ( for Ba (OH)2, α = 1 )

Question 24.
What is the molality of a 10% w/w aqueous sodium hydroxide solution?
(a) 2.778
(b) 2.5
(c) 10
(a) 0.4
Answer:
(b) 2.5
Solution:
100% \(\frac { w }{ w }\) aqueous NaOH solution means that 10 g of sodium hydroxide in 100g solution.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-13

Question 25.
The correct equation for the degree of an associating solute, ‘n’ molecules of which undergoes association in solution, is ………
(a) α = \(\frac { n(i – 1) }{ n – 1 }\)
(b) α2 = \(\frac { n(1 – i) }{ n – 1 }\)
(c) α = \(\frac { n(i – 1) }{ 1 – n }\)
(d) α = \(\frac { n(1 – i) }{ n(1 – i) }\)
Answer:
(c) α = \(\frac { n(i – 1) }{ 1 – n }\)
Solution:
α = \(\frac { (i – 1)n }{ (n – 1) }\) (or) \(\frac { n(i – 1) }{ (1 – n) }\)

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 26.
Which of the following aqueous solutions has the highest boiling point?
(a) 0.1 M KNO3
(b) 0.1 M Na3PO4
(c) 0.1 M BaCl2
(d) 0.1 M K2SO4
Answer:
(a) 0.1 M KNO3
Solution:
Elevation of boiling point is more in the case of Na3PO4(no. of ions 4; 3 Na+, PO43-)

Question 27.
The freezing point depression constant for water is 1.86° k kg mo1-1 . If 5g Na2SO4 is dissolved in 45g water, the depression in freezing point is 3.64°C. The van’t Hoff factor for Na2SO4 is ……..
(a) 2.50
(b) 2.63
(c) 3.64
(d) 5.50
Answer:
(a) 2.50
Solution:
Kf = 1.86
W2 = 5g
∆Tf = 3.64
M2 = 142
W1 = 45g
ΔTf = i x Kf
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-14

Question 28.
Equimolal aqueous solutions of NaCI and KCI are prepared. If the freezing point of NaCI is – 2°C, the freezing point of KCI solution is expected to be ………
(a) – 2°C
(b) – 4°C
(c) – 1°C
(d) 0°C
Answer:
(a) – 2°C
(b) – 4°C
(c) – 1°C
(d) 0°C
Solution:
Equimolal aqueous solution of KCI also shows 2° C depression in freezing point.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 29.
Phenol dimerises in henzene having van’t Hoff factor 0.54. What is the degree of association?
(a) 0.46
(b) 92
(c) 46
(d) 0.92
Answer:
(d) 0.92
Solution:
α = \(\frac { (1-i)n }{ (n-1) }\) = \(\frac { (1 – 0.54)2 }{ (2 – 1) }\) = 0.46 x 2 = 0.92
Question 30.
Assertion: An ideal solution obeys Raoult’s Law
Reason: In an ideal solution, solvent-solvent, as well as solute-solute interactions, are similar to solute-solvent interactions.
(a) both assertion and reason are true and reason is the correct explanation of assertion
(b) both assertion and reason are true but reason is not the correct explanation of assertion
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(a) both assertion and reason are true and reason is the correct explanation of assertion

Samacheer Kalvi 11th Chemistry Solutions Short Answer Questions

Question 31.
Define

  1. Molality
  2. Normality

Answer:
1. Molality (m):
It is defined as the number of moles of the solute present in 1 kg of the solvent
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-15

2. Normality (N):
It is defined as the number of gram equivalents of solute in I litre of the solution.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-16

Question 32.
What is a vapour pressure of liquid? What is relative lowering of vapour pressure?
Answer:
1. The pressure of the vapour in equilibrium with its liquid ¡s called vapour pressure of the liquid at the given temperature.

2. The relative lowering of vapour pressure is defined as the ratio of lowering of vapour. pressure to vapour pressure of pure solvent. Relative lowering of vapour pressure
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-17

Question 33.
State and explain Henry’s law.
Answer:
“The partial pressure of the gas in vapor phase (vapour pressure of the solute) is directly proportional to the mole fraction (x) of the gaseous solute in the solution at low concentrations”. This statement is known as Henry’s law.
Henry’s law can be expressed as,
Psolute α xsolute in solution
Psolute = KH xsolute solution
Here, Psolute represents the partial pressure of the gas in vapour state which is commonly called vapour pressure. Xsolute in solution represents the mole fraction of solute in the solution. KH is an empirical constant with the dimensions of pressure.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 34.
State Raoult law and obtain expression for lowering of apour pressure when the nonvolatile solute is dissolved in solvent.
Answer:
Raoult’s law:
This law states that “in the case of a solution of volatile liquids the partial vapour pressure of each component (A & B) of the solution is directly proportional to its mole fraction.
PA ∝ x A
when xA = 1,
then k = P°A
(P°A = vapour pressure of pure component)
PA = P°A . xa
PB = P°B . xb
when a non-volatile is dissolved in pure water, the vapour pressure of the pure solvent will decrease. In such a solution, the vapours pressure of the solution will depend only on the solvent molecules as the solute is non-volatile.
Psolution ∝ xA
Psolution = k . xA
xA = 1, k = P°solvent
solution = P°solvent – Psolution
Lowering of vapour pressure = P°solvent – Psolution
Relative lowering of vapour pressure = \(\frac { P° – P }{ P° }\) = xB
where xB = Mole fraction of solute.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 35.
What is molal depression constant? Does it depend on nature of the solute?
Answer:
Kf = molar freezing point depression constant or cryoscopic constant.
∆Tf = Kf . m,
where
∆Tf = depression in freezing point.
m = molality of the solution
Kf = cryoscopic constant
If m = I
∆Tf = Kf
i.e., cryoscopic constant is equal to the depression in freezing point for 1 molal solution cryoscopic constant depends on the molar concentration of the solute particles. Kf is directly proportional to the molal concentration of the solute particles.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-18
WB = mass of the solute
WA = mass of solvent
MB = molecular mass of the solute.

Question 36.
What is osmosis?
Answer:
“The phenomenon of the flow of solvent through a semipermeable membrane from pure solvent to the solution is called osmosis”. Osmosis can also be defined as “the excess pressure which must be applied to a solution to prevent the passage of solvent into it through the semipermeable membrane”. Osmotic pressure is the pressure applied to the solution to prevent osmosis.

Question 37.
Define the term isotonic
Answer:
1. Two solutions having same osmotic pressure at a given temperature are called isotonic solutions.

2. When such solutions arc separated by a semipermeable membrane, solvent flow between one to the other on either direction is same. i.e. the net solvent flow between two isotonic solutions is zero.

Samacheer Kalvi 11th Chemistry Solutions Long Answer Questions

Question 38.
You are provided with a solid ‘A’ and three solutions of A dissolved in water – one saturated, one unsaturated, and one supersaturated. How would you determine each solution?
Answer:
1. Saturated solution:
When the maximum amount of solute is dissolved in a solvent, any more addition of solute will result in precipitation at a given temperature and pressure. Such a solution is called a saturated solution.

2. Unsaturated solution:
When the minimum amount of solute is dissolved in a solvent at a given temperature and pressure is called an unsaturated solution.

3. Supersaturated solution:
It is a solution that holds more solute than it normally could in its saturated form.

Example:

  1. A saturated solution where the addition of more compound would not dissolve in the solution. 359 g of NaCI in 1 litre of water at 25°C.
  2. An unsaturated solution has the capacity to dissolve more of the compound. 36 g of NaCI in 1 litre of water at 25°C.
  3. A supersaturated solution is a solution in which crystals can start growing. 500 g of NaCI in 1 litre of water at 25°C.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 39.
Explain the effect of pressure on solubility.
Answer:
Generally, the change in pressure does not have any significant effect on the solubility of solids and liquids as they are not compressible. However, the solubility of gases generally increases with an increase in pressure.

Consider a saturated solution of a gaseous solute dissolved in a liquid solvent in a closed container. In such a system, the following equilibrium exists.
Gas (in a gaseous state) ⇌ Gas (in solution)

According to the Le-Chatelier principle, the increase in pressure will shift the equilibrium in the direction which will reduce the pressure. Therefore, more gaseous molecules dissolve in the solvent, and the solubility increases.

Question 40.
A sample of 12 M Concentrated hydrochloric acid has a density of 1.2 gL-1. Calculate the molality.
Answer:
Given:
Molarity = 12 M HCI
Density of the solution = 1.2 g L-1
In the 12 M HCl solution, there are 12 moles of HCl in 1 litre of the solution.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-19
Calculate mass of water (solvent)
Mass of 1 litre HCI solution = density x volume
= 1.2gmL-1 x 1000 mL
= 1200g
Mass of LICI = No. of moles of HCI x molar mass of HCI
= 12mol x 36.5 g mol-1
= 438g
Mass of waler = mass of HCI solution – mass of HCI
Mass of waler = 1200 – 438 = 762 g
Molalily =\(\frac { 12 }{ 0.762 }\) = 15.75m

Question 41.
A 0.25 M glucose solution at 370.28 K has approximately the pressure as blood. What is the osmotic pressure of blood?
Solution.
C = 0.25 M
T = 370.28 K
(π)glucose = CRT
(π) = 0.25 mol L-1 × 0.082L atm K-1mol-1 × 370.28K
= 7.59 atm

Question 42.
Calculate the molarity of a solution containing 7.5g of glycine (NH2 – CH2 – COOH) dissolved in 500g of water.
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-20

Question 43.
Which solution has the lower freezing point? 10g of methanol (CH3OH) in 100g of water (or) 20g of ethanol (C2H5HO) In 200g of water.
Solution:
∆Tf = Kf . m i.e. ∆Tf ∝ m
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-21
∴ Depression in freezing point is more in methanol solution and it will have a lower freezing point.

Question 44.
How many moles of solute particles are present in one litre of 10-4 M potassium sulphate?
Solution:
In 10-4M K2SO4 solution, there are 10-4 moles of potassium sulphate.
K2SO4 molecule contains 3 ions (2 K+ and 1SO42-)
1 mole of K2SO4 contains 3 x 6.023 x 1023 ions
10 mole of K2SO4 contains 3 x 6.023 x 102 x 10-4 ions = 18.069 x 1019

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 45.
Henry’s law constant for solubility of methane in benzene is 4.2 x 10-5 mm Hg at a particular constant temperature. At this temperature, calculate the solubility of methane at

  1. 750 mm Hg
  2. 840 mmHg

Solution:
(KH)Benzene = 4.2 x 10-5 mm Hg. Solubility of methane = ? P = 750 mm Hg, p = 840 mm Hg
According to Henry’s Law,
P = KH . xsolution
750 mm Hg = 4.2 x 10-5 mm Hg . xsolution
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-24

Question 46.
The observed depression in the freezing point of water for a particular solution is 0.093°C. Calculate the concentration of the solution in molality. Given that the molal depression constant for water is 1.86 K kg mol-1.
Solution:
T1= 0.093°C = 0.093K
m = ?
Kf = 1.86K kg mol-1
∆Tf = kf . m
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-24

Question 47.
The vapour pressure of pure benzene (C6H6) at a given temperature is 640 mm Hg. 2.2 g of non-volatile solute is added to 40 g of benzene. The vapour pressure of the solution is 600 mm Hg. Calculate the molar mass of the solute?
Solution:
P0C6H6 = 640 mm Hg
W2 = 2.2 g (non volatile solute)
W1 = 40 g (benzene)
Psolution = 600 mm Hg
M2 = ?
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-25

Samacheer Kalvi 11th Chemistry Solutions In Text Questions – Evaluate Yourself

Question 1.
If 5.6 g of KOH is present in (a) 500 mL and (b) I litre of solution, calculate the molarity of each of these solutions.
Solution.
Mass of KOH = 5.6g
No. of moles = \(\frac { 5.6 }{ 5.6 }\) = 0.1 mol
1. Volume of the solution = 500 ml = 0.5 L

2. Volume of the solution = IL
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-26

3. Volume of the solution = IL
Molarity = \(\frac { 0.1 }{ 1 }\) M

Question 2.
2.82 g of glucose is dissolved in 30 g of water. Calculate the mole fraction of glucose and water.
Solution:
Mass of glucose = 2.82 g
No. of moles of glucose = \(\frac { 2.82 }{ 180 }\) = 0.0 16
Mass of water = 30g = \(\frac { 30 }{ 18 }\) = 1.67
xH2O = \(\frac { 1.67 }{ 1.67 + 0.016 }\) = \(\frac { 1.67 }{ 1.686 }\) = 0.99
xH2O + xglucose = 1
0.99 + xglucose = 1
xglucose = 1 – 0.99 = 0.01

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 3.
The antiseptic solution of iodopovidone for the use of external application contains 10% w/v of iodopovidone. Calculate the amount of iodopovidone present in a typical dose of 1.5 mL.
Solution:
10% \(\frac { w }{ v }\) means that 10 g of solute in 100 ml solution
∴ Amount of iodopovidone in 1.5 ml = \(\frac { 10g }{ 100ml }\) x 1.5 ml = 0.15 g

Question 4.
A litre of sea water weighing about 1.05 kg contains 5 mg of dissohed oxygen (O2). Express the concentration of dissolved oxygen in ppm.
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-28

Question 5.
Describe how would you prepare the following solution from pure solute and solvent

  1. 1 L of aqueous solution of 1.5 M COCI2.
  2. 500 mL of 6.0 % (v/v) aqueous methanol solution.

Solution:

  1. mass of 1.5 moles of COCI2 = 1.5 x 129.9 = 194.85g
  2. 194.85g anhydrous cobalt chloride is dissolved in water and the solution is make up to one litre in a standard flask.

Question 6.
How much volume of 6 M solution of NaOH is required to prepare 500 mL of 0.250 M NaOH solution.
Solution:
6% \(\frac { v }{ v }\) aqueous solution contains 6g of methanol in 100 ml solution. To prepare 500 ml of 6% v/v solution of methanol 30g methanol is taken in a 500 ml standard flask and required quantity of water is added to make up the solution to 500 ml.

Question 7.
Calculate the proportion of O2 and N2 dissolved in water at 298 K. When air containing 20% O2 and 80% N2 by volume is in equilibrium with water at 1 atm pressure. Henry’s law constants for two gases are KH(O2) = 4.6 x atm and KH (N2) 8.5 x 104 atm.
Solution:
C1V1 = C2V2
6M (V1) = 0.25M x 500 ml
V1 = \(\frac { 0.25 x 500 }{ 6 }\)
V1 = 20.3 mL

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 8.
Explain why the aquatic species are more comfortable in cold water during winter season rather than warm water during the summer.
Solution:
Total pressure = 1 atm
PN2 = \((\frac { 80 }{ 100 })\) x Total pressure = \(\frac { 80 }{ 100 }\) x 1 atm = 0.8 atm
PO2 = \((\frac { 20 }{ 100 })\) x 1 = 0.2 atm
According to Henry’s Law
Psolute = KH x solute in solution
PN2 = (KH)Nitrogen x Mole fraction of Nitrogen in solution
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-29

Question 9.
Calculate the mole fractions of benzene and naphthalene in the vapour phase when an ideal liquid solution is formed by mixing 128 g of naphthalene with 39g of benzene. It is given that the vapour pressure of pure benzene is 50.71 mm Hg and the vapour pressure of pure naphthalene is 32.06 mm Hg at 300 K.
Solution:
P0pure benzene = 50.71 mm Hg
P0nepthalene = 32.06 mm Hg
Number of moles of benzene = \(\frac { 39 }{ 78 }\) = 0.5 mol
Number of moles of naphthalcne = \(\frac { 128 }{ 128 }\) =1 mol
Mole fraction of benzene = \(\frac { 0.5 }{ 1.5 }\) = 0.33
Mole fraction of naphthalene = 1 – 0.33 = 0.67
Partial vapour pressure of benzene =P0benzene x Mole fraction of benzene
= 50.71 x 0.33 = 16.73 mm Hg
Partial vapour pressure of naphthalene = 32.06 x 0.67 = 21.48mm Hg
Mole fraction of benzene in vapour phase = \(\frac { 16.73 }{ 16.73 + 21.48 }\) = \(\frac { 16.73 }{ 38.21 }\) = 0.44
Mole fraction of naphthalene in vapour phase = 1 – 0.44 = 0.56

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 10.
Vapour pressure of a pure liquid A is 10.0 torr at 27°C. The vapour pressure is lowered to 9.0 torr on dissolving one grani of B in 20g of A. If the molar mass of A is 200 g mol-1 then calculate the molar mass of B.
Solution:
P0A = 10 torr
Psolution = 9 torr
WA = 20 g
WB = 1 g
MA = 200 g mol-1
MB = ?
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-30

Question 11.
2.56g of Sulphur is dissolved in 100g of carbon disuiphide. The solution boils at 319.692K. What is the molecular formula ofSulphur in solution? The boiling pointof CS2 is 319. 450K. Given that Kb for CS2 = 2.42 K kg mol-1
Solution:
W2 = 2.56g
W1 = 100g
T = 319.692 K
Kb = 2.42 K kg mol-1
∆Tb = (319.692 – 319.450) K = 0.242 K
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-31
M2 = 256g mol-1
Molecular mass of sulphur in solulion = 256 g mol-1
Atomic mass of one mole of sulphur atom = 32
No. of atoms in a molecule of sulphur = \(\frac { 256 }{ 32 }\) = 8
Hence, molecular tòrmula of sulphur is S8.

Question 12.
2g of a non-electrolyte solute dissolved in 75g of benzene lowered the freezing point of benzene by 0.20 K. The freezing point depression constant of benzene is 5.12 K Kg mol-1. Find the molar mass of the solute.
Solution:
W2 = 2g
W1 = 75g
∆Tf = 0.2 K
kf = 5.12 K kg mol-1
M2 = ?
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-32

Question 13.
What is the mass of glucose (C6H12O6) in its one-litre solution is isotonic with 6g L-1 of urea (NH2CONH2)?
Solution:
The osmotic pressure of urea solution (π1) = CRT
\(\frac { { W }_{ 2 } }{ { M }_{ 2 }V }\)RT = \(\frac { 6 }{ 60 x 1 }\) x RT
The osmotic pressure of glucose solution
2) \(\frac { { W }_{ 2 } }{ 180\times 1 }\) x RT
For isotonic solution, π1 = π2
\(\frac { 6 }{ 60 }\) = \(\frac { { W }_{ 2 } }{ 180\times 1 }\) RT
⇒ W2 = \(\frac { 6 }{ 60 }\) x 180
⇒ W2 = 18 g

Question 14.
0.2m aqueous solution of KCI freezes at – 0.68°C calculate van’t Hoff factor. Kf for water is 1.86 K kg mol-1.
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-36
Given,
∆Tf = 0.680 K
m = 0.2 m,
∆Tf (observed) = 0.680K
∆Tf (Calculated) = kf
m = 1.86 K kg mol-1 x 0.2 mol kg-1 = 0.372K
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-37

Samacheer Kalvi 11th Chemistry Solutions Example problems Solved

Question 1.
What volume of 4M HCI and 2M HCI should be mixed to get 500 mL of 2.SM HCI?
Solution:
Let the volume of 4M HCl required to prepare 500 mL of 2.5 M HCI = x mL
Therefore, the required volume of 2M HCI = (500 – x) mL
We know from the equation x = \(\frac { 250 }{ 2 }\) = 125 mL
Hence, volume of 4M HCI required = 125 mL
Volume of 2M HCl required = (500 – 125) mL = 375 mL

Question 2.
0.24g of a gas dissolves in 1 L of water at 1.5 atm pressure. Calculate the amount of dissolved gas when the pressure is raised to 6.0 atm at constant temperature.
Solution:
Psolute = KH xsolute in solution
At pressure 1.5 atm, p1 = KH x1 ………..(1)
At pressure 6.0 atm, p2 = KHx2 …………..(2)
Dividing equation (1) by (2)
Weget
\(\frac { { P }_{ 1 } }{ { P }_{ 2 } }\) = \(\frac { { x }_{ 1 } }{ { x }_{ 2 } }\)
\(\frac { 1.5 }{ 6.0 }\) = \(\frac { { 0.24 } }{ { x }_{ 2 } }\)
Therefore
x2 = \(\frac { 0.24×6.0 }{ 1.5 }\) = 0.96 g/L

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 3.
An aqueous solution of 2% nonvolatile solute exerts a pressure of 1.004 bar at the boiling point of the solvent. What is the molar mass of the solute when PA° is 1.013 bar?
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-38
In a 2% solution weight of the solute is 2g and solvent is 98g
ΔP = PA0 – Psolution = 1.013 – 1.004 bar = 0.009 bar
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-39

Question 4.
0.75 g of an unknown substance is dissolved in 200 g solvent. If the elevation of boiling point is 0.15 K and molal elevation constant is 7.5K kg more then, calculate the molar mass of unknown substance.
Solution:
∆Tb = Kb m = Kb x W2 x 1000/M2 x W1
M2 = Kb x W2 x 1000/∆Tb x W1
= 7.5 x 0.75 x 1000/0.15 x 200 = 187.5g mol-1

Question 5.
Ethylene glycol (C2H6O2) can be used as an antifreeze in the radiator of a car. Calculate the temperature when ice will begin to separate from a mixture with 20 mass percent of glycol in water used in the car radiator. Kf for water = 1.86 K kg mol-1 and molar mass of ethylene glycol is 62g mol-1.
Solution:
Weight of solute (W2) = 20 mass percent of solution means 20g of ethylene glycol
Weight of solvent (water) W1 = 100 – 20 = 80g
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-40
The temperature at which the ice will begin to separate is the freezing of water after the addition of solute i.e. 7.5 K lower than the normal freezing point of water (273 – 7.5)K = 265.5K

Question 6.
At 400K 1.5 g of an unknown substance is dissolved in solvent and the solution is made to 1.5 L. Its osmotic pressure is found to be 0.3 bar. Calculate the molar mass of the unknown substance.
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-41

Question 7.
The depression in freezing point is 0.24K obtained by dissolving 1g NaCI in 200g water. Calculate van’t – Hoff factor. The molal depression constant is 1.86 K kg mol-1.
Solution:
Sol. Molar mass of solute
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-42

Samacheer Kalvi 11th Chemistry Solutions Additional Questions Solved

Samacheer Kalvi 11th Chemistry Solutions 1 Mark Questions and Answers

I. Choose the correct answer.

Question 1.
Among the following, which one is mostly present in seawater?
(a) NaCI
(b) Nal
(c) KCI
(d) MgBr2
Answer:
(a) NaCI

Question 2.
Statement I: The most common property of seawater and air is homogeneity.
Statement II: The homogeneity implies uniform distribution of their constituents through the mixture.
(a) Statements I and II arc correct and II are the correct explanation of I.
(b) Statements I and II are correct but II is not the correct explanation of I.
(c) Statement I is correct but II is wrong.
(d) Statement I is wrong but II is correct.
Answer:
(a) statement I and II are correct and II is the correct explanation I.

Question 3.
Which one of the following is a homogeneous mixture?
(a) Seawater
(b) Air
(c) Alloys
(d) All the above
Answer:
(d) All the above

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 4.
Statement I: The salt solution is an aqueous solution.
Statement II: If water is used as the solvent, the resultant solution is called an aqueous solution.
(a) Statements I and II are correct but II is not the correct explanation of I.
(b) Statements I and II are correct and II is the correct explanation of I.
(c) Statement I is correct but statement II is wrong.
(d) Statement I is wrong but statement II is correct.
Answer:
(b) Statements I and II are correct and II is the correct explanation of I.

Question 5.
Statement I: The dissolution of ammonium nitrate increases steeply with an increase in temperature.
Statement II: The dissolution process of ammonium nitrate is endothermic in nature.
(a) Statement I and II are correct and statement II is the correct explanation of statement I.
(b) Statement I and II are correct but II is not the correct explanation of I.
(c) Statement I is correct but II is wrong.
(d) Statement I is wrong but II is correct.
Answer:
(a) Statement I and II are correct and statement II is the correct explanation of statement I.

Question 6.
In which of the following compound the solubility decreases with the increase of temperature?
(a) sodium chloride
(b) ammonium nitrate
(c) ceric sulphate
(d) calcium chloride
Answer:
(c) ceric sulphate

Question 7.
Which of the following is not an ideal solution?
(a) Benzene & toluene
(b) n – Hexane & n – Heptane
(c) Ethyliodide & ethyl bromide
(d) Ethanol and water
Answer:
(d) Ethanol and water

Question 8.
Which one of the following shows positive deviation from Raoult’s law?
(a) Ethyliodide and Ethyl bromide
(b) Ethyl alcohol and cyclohexane
(c) Chioro benzene & bromo benzene
(d) Benzene & toluenc
Answer:
(b) Ethyl alcohol and cyclohexane

Question 9.
Which one of the following is not a non-ideal solution showing positive deviation?
(a) Benzene & acetone
(b) CCl4 & CHCl3
(c) Acetone & ethyl alcohol
(d) Benzene and toluene
Answer:
(d) Benzene and toluene

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 10.
Which of the following shows a negative deviation from Raoult’s law?
(a) Phenol and aniline
(b) Benzene and toluene
(c) Acetone and ethanol
(d) Bcnzene and acetone
Answer:
(a) Phenol and aniline

Question 11.
Which of the following is not a non-ideal solution showing negative deviation?
(a) Phenol and aniline
(b) Ethanol and water
(c) Acetone + Chloroform
(d) n – Heptane and n – Hexane
Answer:
(d) n – Heptane and n-Hexane

Question 12.
Statement I: A solution of potassium chloride in water deviates from ideal behavior.
Statement II: The solute dissociates to give K and Cl ions which form strong ion-dipole interaction with water molecules.
(a) Statement I & II are correct and II is the correct explanation of I
(b) Statement I & II are correct but II is not the correct explanation of I
(c) Statement I is correct but statement II is wrong.
(d) Statement I is wrong but statement II is correct.
Answer:
(a) Statement I & II are correct and II is the correct explanation of I

Question 13.
Statement I: Acetic acid deviates from ideal behavior.
Statement II: Acetic acid exists as a dimer by forming intermolecular hence deviates from Raoult’s law.
(a) Statement I & II are correct and II is the correct explanation of I.
(b) Statement I & II are correct but II is not the correct explanation of I.
(c) Statement I is true but II is wrong.
(d) Statement I is wrong but II is correct.
Answer:
(a) Statement I & II are correct but II is the correct explanation of I.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 14.
Which one of the following has found to have abnormal molar mass? hydrogen bonds and
(a) NaCl
(b) KCI
(c) Acetic acid
(d) all the above
Answer:
(d) All the above

Question 15.
What would be the value of the van’t Hoff factor for a dilute solution of K2SO4 in water?
(a) 3
(b) 2
(c) 1
(d) 4
Answer:
(a) 3
Solution:
ions produced = n = 3
Since
K2SO4 → 2K+ + SO42-
K2SO4 is completely dissociated so
∝ = \(\frac { i – 1 }{ n – 1 }\) = \(\frac { i – 1 }{ 3 – 1 }\) = 1
i – 1 = 1 x 2
i – 1 = 2
i = 2+1 = 3

Question 16.
In the determination of the molar mass of AB using a colligative property, what may be the value of van’t Hoff factor if the solute is 50% dissociates?
(a) 0.5
(b) 1.5
(c) 2.5
(d) 1
Answer:
(b) 1.5
Solution:
∝ = \(\frac { i – 1 }{ n – 1 }\) = 0.5
\(\frac { i – 1 }{ 2 – 1 }\) = 0.5
i – 1 = 0.5
i = 0.5 + 1 = 1.5

Question 17.
Which of the following solution has the highest boiling point?
(a) 5.85% solution of NaCI
(b) 18.0% solution of glucose
(c) 6.0% solution of urea
(d) All have the same boiling point
Answer:
(a) 5.85% solution of NaCl

Question 18.
Which one of the following pair is called an ideal solution?
(a) nicotine – water
(b) water – ether
(c) water – alcohol
(d) Chiorobenzene – bromobenzene
Answer:
(d) Chiorobenzene – bromobenzene

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 19.
Which of the following is not a colligative property?
(a) optical activity
(b) osmotic pressure
(c) elevation boiling point
(d) depression in freezing point
Answer:
(a) optical activity

Question 20.
On dissolving sugar in water at room temperature solution feels cool to touch. Under which of the following cases dissolution of sugar will be most rapid?
(a) Sugar crystals in cold water
(b) Sugar crystals in hot water
(c) powdered sugar in cold water
(d) powdered sugar in hot water
Answer:
(d) powdered sugar in hot water

II. Match the following.
Question 1.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-43
Answer:
(a) 3 4 1 2

Question 2.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-44
Answer:
(d) 3 4 2 1

Question 3.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-45
Answer:
(c) 2 4 1 3

Question 4.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-46
Answer:
(a) 4 3 1 2

Question 5.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-47
Answer:
(b) 2 4 1 3

III. Fill in the blanks.

Question 1.
……… covers more than 70% of the earth’s surface.
Answer:
Seawater

Question 2.
……… is an important naturally occurring solution.
Answer:
Air

Question 3.
An example of solid homogeneous mixture is ……….
Answer:
Brass

Question 4.
A mixture of N2, O2, CO2 and other traces of gases is known as ………
Answer:
Air

Question 5
……… a non – aqueous solution.
Answer:
Br2 in CCl4

Question 6.
……… is an example of a gaseous solution.
Answer:
Camphor in nitrogen gas

Question 7.
……… is used for a dental filling.
Answer:
Amalgam of potassium

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 8.
Carbonated water is an example for ………
Answer:
Liquid solution

Question 9.
Humid oxygen is an example of ………
Answer:
Gaseous solution

Question 10.
The concentration of commercially available H2O2 is ………
Answer:
3%

Question 11.
The molality of the solution containing 45g of glucose dissolved in 2kg of water is ………
Answer:
0.125m
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-48

Question 12.
5.845 g of NaCl is dissolved in water and the solution was made up to 500 mL using a standard flask. The strength of the solution in molarity is ………
Answer:
0.2 M
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-49

Question 13.
3.15 g of oxalic acid dihydrate is dissolved in water and the solution was made up to 100 ml using a standard flask. The strength of the solution in normality is ………
Answer:
0.5N
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-50

Question 14.
5.85 g of NaCl is dissolved in water and the solution was made upto 500 ml using a standard flask. The strength of the solution in formality is ………
Answer:
0.2 F
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-51

Question 15.
Neomycin, aminoglycoside antibiotic cream contains 300 mg of neomycin sulphate the active ingredient in 30 g of ointment base. The mass percentage of neomycin is ………
Answer:
1%
Solution:
The mass percentage of neomycin
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-52

Question 16.
0.5 mole of ethanol is mixed with 1.5 moles of water. Then the mole fraction of ethanol and water are ……….
Answer:
0.25, 0.75
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-53
= \(\frac { 0.5 }{ 1.5 + 0.5 }\) = \(\frac { 0.5 }{ 2.0 }\) = 0.25
Mole fraction of water = \(\frac { 1.5 }{ 2.0 }\) = 0.75

Question 17.
50 mL of tincture of benzoin, an antiseptic solution contains 10 ml of benzoin. The volume percentage of benzoin is ……….
Answer:
20%
Solution:
Volume percentage of benzoin
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-54
= \(\frac { 10 }{ 50 }\) x 100 = 20%

Question 18.
A 60 ml of paracetamol pediatric oral suspension contains 3g of paracetamol. The mass percentage of paracetamol is …………
Answer:
5%
Solution:
Mass percentage of paracetamol =
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-55
= \(\frac { 3 }{ 60 }\) x 100 = 5%

Question 19.
50 ml of tap water contains 20 mg of dissolved solids. The TDS value in ppm is ………..
Answer:
400 ppm
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-56

Question 20.
The concentration term used in the neutralisation reactions is …………
Answer:
Normality

Question 21.
The concentration term is used in the calculation of vapour pressure of the solution is …………..
Answer:
Mole fraction

Question 22.
The term used to express the active ingredients present in therapeutics is ………
Answer:
Percentage units

Question 23.
When the maximum amount of solute is dissolved in a solvent at a given temperature, the solution is called ………..
Answer:
Saturated solution

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 24.
The solvent in which sodium chloride readily dissolves is …………
Answer:
Water

Question 25.
………… is used by deep-sea divers.
Answer:
Helium, nitrogen and oxygen

Question 26.
The mathematical expression of Raoult’s law is ………..
Answer:
PA = PA0 . XA

Question 27.
……….. is an ideal solution?
Answer:
Chloro benzene & bromo benzene

Question 28.
………….. is important in some vital biological systems.
Answer:
osmotic pressure

Question 29.
………. is not a colligative property.
Answer:
vapour pressure

Question 30.
According to van’t Hoff equation. the value of osmotic pressure t is equal to …………
Answer:
π = CRT

Question 31.
The osmotic pressure of the blood cells is approximately equal to 37°C.
Answer:
7 atm.

Question 32.
Which one of the following is applied in water purification?
Answer:
reverse osmosis

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 33.
In the commercial reverse osmosis process, the semi-permeable membrane used is ………..
Answer:
cellulose acetate

Question 34.
The degree of dissociation α is equal to ……….
Answer:
\(\frac { i – 1 }{ n – 1 }\)

Question 35.
The degree of association a is equal to ……….
Answer:
\(\frac { (i – 1)n }{ n – 1 }\)

Question 36.
The estimated vantt Hoff factor for acetic acid solution in benzene is ………..
Answer:
0.5

Question 37.
The estimated van’t Hoff factor for sodium chloride in water is ………..
Answer:
2

Question 38.
Number of moles of the solute dissolved per dm3 of solution is ……….
Answer:
molarity

Question 39.
Molarity of pure water is ………….
Answer:
55.55
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-57
Question 40.
18 g of glucose is dissolved in 90 g of water. The relative lowering of vapour pressure is equal to ………..
Answer:
0.1
Solution:
\(\frac { P° – P }{ P° }\) = x2
x2 = No. of moles of glucose
\(\frac { 18 }{ 180 }\) = 0.1
\(\frac { P° – P }{ P° }\) = 0.1

Question 41.
When NaCl is dissolved in water, boiling point ………..
Answer:
increases

Question 42.
Use of glycol as antifreeze in an automobile is an important application of …………….
Answer:
Colligative property

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 43.
Ethylene glycol is mixed with water and used as antifreeze in radiators because …………..
Answer:
it lowers the freezing point of water

Question 44.
The colligative properties of a solution depend on ………… present in it.
Answer:
Number of solute particles

Question 45.
Low concentration of oxygen in the blood and tissues of people living at high altitude is due to ………….
Answer:
low atmospheric pressure

IV. Choose the odd one out.

Question 1.
(a) Air
(b) Camphor in nitrogen gas
(c) Humid oxygen
(d) Saltwater
Answer:
(d) Saltwater.
a, b and e are gaseous solutions whereas d is a liquid solution.

Question 2.
(a) CO2 dissolve in water
(b) Saltwater
(c) Solution of H2 in palladium
(d) Ethanol dissolved in water
Answer:
(c) Solution of H2 in palladium
a, b and d are liquid solutions whereas c is a solid solution.

Question 3.
(a) Amalgam of potassium
(b) Camphor in nitrogen gas
(c) Solution of H2 in palladium
(d) Gold alloy
Answer:
(b) Camphor in nitrogen gas
a, b and d arc solid solutions whereas b is the gaseous solution.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 4.
(a) Vapour pressure
(b) Lowering of vapour pressure
(c) Osmotic pressure
(d) Elevation of boiling point
Answer:
(a) Vapour pressure
b, e and dare colligative properties whereas a is a physical property.

Question 5.
(a) Benzene and tolucne
(b) Chlorobenzene and Bromobenzene
(c) Benzene and acetone
(d) n – hexane and n – heptane
Answer:
(a) Benzene and acetone
a, b, and d are ideal solutions whereas c is a non-ideal solution.

Question 6.
(a) Ethyl alcohol and cyclohexane
(b) Ethyl bromide and ethyl iodide
(c) Acetone and ethyl alcohol
(d) Benzene and acetone
Answer:
(a) Ethyl bromide and ethyl iodide
a, e and dare non-ideal solutions whereas b is an ideal solution.

V. Choose the correct pair.

Question 1.
(a) Humid oxygen – Liquid solution
(b) Gold alloy – Solid solution
(c) Saltwater – Gaseous solution
(d) Solution of H2 in palladium – Gaseous solution
Answer:
(b) Gold alloy – Solid solution

Question 2.
(a) Air – Gaseous solution
(b) Amalgam of potassium – Liquid solution
(c) Saltwater – Solid solution
(d) Carbonated water – Solid solution
Answer:
(a) Air – Gaseous solution

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 3.
(a) Benzene and toluene – Non-ideal solution
(b) Benzcnc and acetone – Non-ideal solution
(c) Chlorobenzene and bromo henzene – Non-ideal solution
(d) Carbon tetrachloride and Chloroform – the ideal solution
Answer:
(b) Benzene and acetone – Non-ideal solution

Question 4.
(a) Benzene and toluene – Ideal solution
(b) n-hexane and n-heptane – Non-ideal solution
(c) Ethyl iodide and ethyl bromide – Non-ideal solution
(d) Chiorobenzene and bromo benzene – Non-ideal solution
Answer:
(a) Benzene and toluene – Ideal solution

VI. Choose the incorrect pair.
Question 1.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-59
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-60

Question 2.
(a) Benzene and acetone – Ideal solution
(b) Ethyl alcohol and cyclohexane – Non-ideal solution
(C) n-hexane and n-heptane – Ideal solution
(d) Chioro benzene – Ideal solution
Answer:
(a) Benzene and acetone – Ideal solution

VII Assertion & Reason.

Question 1.
Assertion (A): When NaCI is added to water, depression in the freezing point is observed.
Reason (R): The lowering of vapour pressure of a solution causes the depression in freezing point.
(a) Assertion and Reason are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Assertion and Reason are correct and R is the correct explanation of A.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 2.
Assertion (A): Ammonia reacts with water does not obey Henry’s law.
Reason (R): The gases reacting with the solvent does not obey Henry’s law.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(c) (A) is correct hut (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 3.
Assertion (A): Acetic acid solution deviates from Raoult’s law.
Reason (R): Association of solute molecules exists as a dimer by forming intermolecular. hydrogen bonds and hence deviates from Raoult’s law.
(a) Both (A) and (R) arc wrong.
(b) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct and (R) is the correct explanation of (A).

VIII. Choose the correct statement.

Question 1.
(a) Raoult’s law is applicable to volatile solid solute in liquid solvent
(b) Henry’s law is applicable to solution containing solid solute in liquid solvent
(c) For very dilute solutions, the solvent obeys Raoult’s law and the solute obeys Henry’s law.
(d) For saturated solution containing volatile solid solute in liquid solvent both laws are obeyed.
Answer:
(c) For very dilute solutions. the solvent obeys Raoult’s law and the solute obeys LIenrys law.

Samacheer Kalvi 11th Chemistry Solutions 2 Marks Questions and Answers

I. Write brief answer to the following questions:

Question 1.
Define solution.
Answer:
A solution is a homogeneous mixture of two or more substances, consisting of atoms, ions or molecules. The constituent of the homogeneous mixture present in a lower amount is called the solute, and the one present in a larger amount is called the solvent. For example, when a small amount of NaCl dissolved in water.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 2.
Define the solution with an example.
Answer:
1. A solution is a homogeneous mixture of two or more substances consisting of atoms. ions or molecules.

2. For example, when a small amount of NaCl is dissolved in water, a homogeneous solution is obtained. In this solution, Na+ and C ions are uniformly distributed in the water. Here NaCI is the solute and water is the solvent.

Question 3.
What is an unsaturated solution?
Answer:
An unsaturated solution is one that contains less amount of solute than its capacity to dissolve.

Question 4.
Define molality.
Answer:
Molality is defined as the number of moles of solute present in 1 kg of the solvent.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-61

Question 5.
Define molarity.
Answer:
Molarity is defined as the number of moles of solute present in 1 litre of the solution.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-62

Question 6.
Define normality.
Answer:
Normality is defined as the number of gram equivalents of solute present in 1 litre of the solution.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-63

Question 7.
Define formality.
Answer:
Formality (F) is defined as the number of formula weight of solute present in 1 litre of the solution.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-64

Question 8.
Define mole fraction.
Answer:
The mole fraction of a component is the ratio of a number of moles of the component to the total number of moles of all components present in the solution.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-65

Question 9.
Show that the sum of mole fraction of a solution is equal to one.
Answer:
Consider a solution containing two components A and 13 whose mole fractions are xA and xB respectively. Let the number of moles of two components A and B are nA and nB respectively.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-66

Question 10.
Define mass percentage.
Answer:
Mass percentage is defined as the ratio of the mass of the solute in g to the mass of solution in g multiplied by 100.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-67

Question 11.
Define volume percentage.
Answer:
Volume percentage is defined as the ratio of volume of solute in mL to the volume of solution in ml multiplied by 100.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-68

Question 12.
Define mass by volume percentage.
Answer:
It is defined as the ratio of the mass of the solute in g to the volume of the solution in ml multiplied by 100.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-69

Question 13.
What is meant by ppm? Where is it used?
Answer:
1. part per million =
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-70

2. ppm is used to express the quantity of solutes present in small amounts in solutions.

Question 14.
What is the elevation of boiling point? Give it.
Answer:
The temperature difference between the solution and pure solvent is called elevation of boiling point,
∆Tb = T – T°
Unit of ∆Tb is K Kg mole-1.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 15.
Define solubility.
Answer:
The solubility of a substance is defined as the amount of the solute that can be dissolved in loo g of the solvent at a given temperature to form a saturated solution.

Question 16.
Ammonia is more soluble than oxygen in water. Why?
Answer:
Ammonia forms hydrogen bonding with water molecules, this intermolecular bonds arc very strong and thus the ammonia is more soluble in water. Ammonia is strongly interact with water to form ammonium hydroxide. But oxygen is more electronegative it is not able to interact with water more. So NH3 is more soluble than O2 in water.

Question 17.
Solubility of a solid solute in a liquid solvent increases with increase in temperature. Justify this statement.
Answer:
When the temperature is increased,the average kinetic energy of the molecules of the solute and the solvent increases. The increase in the kinetic energy facilitates the solvent molecules to break the intermolecular attractive forces that keep the solute molecules together and hence the solubility increases.

Question 18.
Dissolution of ammonium nitrate increases with increase in temperature. Why?
Answer:
The dissolution process of ammonium nitrate is endothermic. So the solubility increases with increase in temperature.

Question 19.
What is the relationship between the solubility of eerie sulphate with temperature?
Answer:
The dissolution of eerie sulphate is exothermic and the solubility decreases with the increase in temperature.

Question 20.
Why in the dissolution of CaCl2, the solubilit increases moderately with high temperature?
Answer:
Even though the dissolution of CaCI2, is cxothcrmic, the soluhility increases moderately with increase in temperature. Here the entropy factor plays a significant role in deciding the position of equilibrium.

Question 21.
Why the carbonated drinks are stored in pressurized container?
Answer:
1. The carbonated beverages contain CO2 dissolved in them. To dissolve the CO2 in these drinks, CO2 gas is bubbled through them under high pressure.

2. These containers are sealed to maintain the pressure. When we open these containers at atmospheric pressure, the pressure of the CO2 drops to the atmospheric pressure level and hence bubbles of CO2 rapidly escape from the solution and show effervescence.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 22.
Define

  1. Evaporation
  2. Condensation.

Answer:
1. Evaporation:
If the kinetic energy of molecules in the liquid state overcomes the intermolecular force of attraction between them, then the molecules will escape from the liquid state. This process is called evaporation.

2. Condensation:
The vapour molecules are in random motion during which they collide with each other and also with the walls of the container. As the collision is inelastic, they lose their energy and as a result, the vapour returns back to a liquid state. This process is called condensation.

Question 23.
State Dalton’s law of partial pressure.
Answer:
According to Dalton’s law of partial pressure, the total pressure in a closed vessel will be equal to the sum of the partial pressure of the individual components.
Ptotal = PA + PB

Question 24.
Give the reason behind the lowering of vapour pressure in the dissolution of NaCl in water?
Answer:
NaCI is a non volatile solute. When a non-volatile solute is dissolved in a pure solvent, the vapour pressure of pure solvent will decrease. In such a solution, vapour pressure of the solution will depend only on the solvent molecules as the solute is non-volatile.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 25.
What is the ideal solution? Give example.
Answer:
An ideal solution is a solution in which each component i.e., the solute as well as the solvent obeys Raoult’s law over the entire range of concentration.

Question 26.
What is volume percentage?
Answer:
It is defined as the volume of the component present in 100mL of the solution. If Va is a volume of solute and Vb is the volume of solvent, then
Vollume percentage of solute = \(\frac{V_{a} \times 100}{V_{a}+V_{b}}\)

Question 27.
What are colligative properties? Give example.
Answer:
The properties which do not depend on the chemical nature of the solute but depend only on the number of solute particles present in the solution are called colligative properties. e.g.,

  1. Relative lowering of vapour pressure – \(\frac { P° – P}{ P° }\)
  2. Osmotic pressure – π
  3. Elevation of the boiling point – ∆Tb
  4. Depression in freezing point – ∆Tf

Question 28.
What is meant by elevation of boiling point?
Answer:
1. The boiling point of a liquid is the temperature at which its vapour pressure becomes equal to the atmospheric pressure.

2. When a non-volatile solute is added to pure solvent at its boiling point, the vapour pressure of the solution is lowered below 1 atm. To bring the vapour pressure again to 1 atm, the temperature of the solution has to be increased.

3. As a result, the solution boils at a higher temperature (Tb) than the boiling point of pure solvent (Tb°). This increase in the boiling point is known as the elevation of boiling point.

Question 29.
Define ebullioscopic constant.
Answer:
Ebullioscopic constant kb, is equal to the elevation in boiling point for 1 molal solution.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-71

Question 30.
What is Van’t Hoff factor?
Answer:
It is defined as the ratio of the actual molar mass to the abnormal (calculated) molar mass of the solute. Here, the abnormal molar mass is the molar mass calculated using the experimentally determined colligative property.
i = \(\frac{\text { Normal (actual) molar mass }}{\text { obseved (abnormal) molar mass }}\)

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 31.
Write the Van’t Hoff equation of osmotic pressure.
Answer:
Van’t Hoff equation states that for dilute solutions, the osmotic pressure is directly proportional to the molar concentration of the solute and the temperature of the solution.
π = CRT
where
π = Osmotic pressure
C = concentration
T = Temperature
R = gas constant

Question 32.
Define Van’t Hoff factor.
Answer:
van’t Hoff factor (I) is defined as the ratin of the actual molar mass to the abnormal molar mass of the solute.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-72

Question 33.
How is degree of dissociation and degree of association are related with van’t Hoff factor?
Answer:
The degree of dissociation or association can be related to van’t Hoff factor
1. using the following relationship

  • αdissociation = \(\frac { i – 1 }{ n – 1 }\)
  • αassociation = \(\frac { (1 – i)n }{ n – 1 }\)

where n = number of solute particles

Question 34.
Give an example of a solid solution ¡n which the solute is a gas.
Answer:
Solution of hydrogen in palladium.

Question 35.
Why the carbonated drinks are stored in a pressurized container?
Answer:
The carbonated beverages contain carbon dioxide dissolved in them. To dissolve the carbon dioxide in these drinks, the CO2 gas is bubbled through them under high pressure. These containers are sealed to maintain the pressure. When we open these containers at atmospheric pressure, the pressure of the CO2 drops to the atmospheric level, and hence bubbles of CO2 rapidly escape from the solution and show effervescence. The burst of bubbles is even more noticeable if the soda bottle in warm condition.

Question 36.
Why do gases always tend to be less soluble in liquids as the temperature is raised?
Answer:
Mostly dissolution of gases in liquid is an exothermic process. it is because of the fact that this process involves a decrease of entropy. Thus, an increase in temperature tends to push the equilibrium towards a backward direction as a result of which solubility of the gas decrease with a rise in temperature.
(Gas + Solvent \(\rightleftharpoons\) Solution + Heat)

Question 37.
Why is the freezing point depression of 0.1 M NaCl solution nearly twice that of 0.1M glucose solution?
Answer:
NaCl is an electrolyte and it dissociates completely whereas glucose being a non-electrolyte does not dissociate. Hence, the number of particles in 0.1 M NaCl solution is nearly double for NaCI solution than that for glucose solution of same molarity.

Therefore depression in freezing point being a colligative property ¡s nearly twice for NaCl solution than that for glucose solution of same molarity.

Question 38.
Why a person suffering from high blood pressure is advised to take a minimum quantity of common salt?
Answer:
Osmotic pressure is directly proportional to the concentration of solutes. Our body fluid contains a number of solutes. On taking large amount of salt, ions entering into the body fluid thereby raises the concentration of solutes. As a result, osmotic pressure increases which may rupture the blood cells.

Samacheer Kalvi 11th Chemistry Solutions 3 Marks Questions and Answers

Question 1.
What are gaseous solution? Give its various types with example
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-73

Question 2.
What are liquid solutions ? Explain with example
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-74

Question 3.
What are solid solution? Give example.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-75

Question 4.
How will you prepare a standard solution?
Answer:

  1. A standard solution or a stock solution is a solution whose concentration is accurately known.
  2. A standard solution of required concentration can be prepared by dissolving a required amount of a solute in a suitable amount of solvent.
  3. It is done by transforming a known amount of solute to a standard flask of definite volume. A small amount of water is added lo the flask and shaken well to dissolve the salt.
  4. Then water is added to the flask to bring the solution level lo the mark indicated at the top end of the flask.
  5. The flask is stoppered and shaken well to make concentration uniform.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 5.
What are the advantages of standard solution.
Answer:
1. The error due to weighing the solute can be minimised by using concentrated stock solution that requires large quantities of solute.

2. We can prepare working standards of different concentrations by diluting the stock solution which is more efficient since consistency is maintained.

3. Some of the concentrated solutions are more stable and are less likely to support microbial growth than working standards used in the experiments.

Question 6.
Explain the solubilities of ammonium nitrate, calcium chloride, ceric sulphate and sodium chloride in water at different temperature with a graph.
Answer:
1. The solubility of sodium chloride does not vary appreciably as the maximum solubility is achieved at normal temperature. In fact, there is only 10% increase in solubility between 0°C to 100°C.

2. The dissolution process of ammonium nitrate is endothermic, the solubility increases with

3. In the case of eerie sulphate. the dissolution is exothermic and the solubility decreases with increase in temperature.

4. Even though the dissolution of calcium chloride is exothermic, the solubility increases moderately with increase in temperature. Here the entropy factor also plays a significant role in deciding the position of equilibrium.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-76

Question 7.
Explain the effect of temperature gaseous solute ¡n liquid solvent.
Answer:
1. In the case of gaseous solute in liquid solvent, the solubility decreases with increase in temperature.

2. When a gaseous solute dissolves in a liquid solvent, its molecules interact with solvent molecules with weak inter molecular forces when the temperature increases, the average. kinetic energy of the molecules present in the solution also increases.

3. The increase in kinetic energy breaks (he weak inter molecular forces between the gaseous solute and liquid solvent with results in the release of the dissolved gas molecules to gaseous state.

4. The dissolution of most of the gases in Liquid solvents is an endothermic process, the increase in temperature decreases the dissolution of gaseous molecules.

Question 8.
Give reason why aquatic species are less sustained in hot water?
Answer:
There will be decrease in solubility of gases in solution with increase in temperature. During summer, in hot water rivers, due to high temperature. the availability of dissolved oxygen decreases. So the aquatic species are less sustained in hot water.

Question 9.
Deep – sea divers use air diluted with helium gas in their tanks. Why? (or) Justify this statement.
Answer:
1. Deep-sea divers carry a compressed air tank for breathing at high pressure under water. This air tank contains nitrogen and oxygen which are not very soluble in blood and other body fluids at normal pressure.

2. As the pressure at the depth is far greater than the surface atmospheric pressure, more nitrogen dissolves in the blood when the diver breathes from tank.

3. When the divers ascends to the surface, the pressure decreases, the dissolved nitrogen comes out of the blood quickly forming bubbles in the blood stream.

These bubbles restrict blood flow, affect the transmission of nerve impulses and can even burst the capillaries or block them. This condition is called “the bends” which are painful and dangerous to life.

4. To avoid such dangerous condition they use air diluted with helium gas (11.7 % helium, 56.2% nitrogen and 32.1% oxygen) of lower solubility of helium in the blood than nitrogen.

Question 10.
What are the limitations of Henry’s law?
Answer:

  1. Henry’s law is applicable at moderate temperature and pressure only.
  2. Only the less soLuble gases obey Henry’s law.
  3. The gases reacting with solvent do not obey Henry’s law.
  4. The gases obeying Henrys law should not be associated or dissociated while dissolving in the solvent.

Question 11.
Explain how benzene in toluene obeys Raoult’s law.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-77
Answer:
The variation of vapour pressure of pure benzene and toluenc with its mole fraction is given in the graph.
1. The vapour pressure of pure toluene and pure benzene are 22.3 and 74.7 mm Hg respectively.

2. The graph shows the partial vapour pressure of pure components increases linearly with the increase of the mole fraction of the respective components. The total pressure at any composition of the solute and solvent is given by the straight line.

3. Psolution = P0toluene + xbenzene (P0benzene – P0toluene)

Question 12.
Derive the relationship between the relative lowering of vapour pressure and mole fraction of the solute.
Answer:
Psolution ∝ xA by Raoult’s law. where xA is the mole fraction of the solvent.
Psolution = k . xA
When
xA = 1
k = P0solvent
P0solvent = partial pressure of pure solvent
Psolution = P0solvent . xA
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-78
where
xB = mole fraction of the solute
xA + xB = 1
xB = 1 – xA
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-79

Question 13.
The depression in freezing point is 0.24 K obtained by dissolving 1 g NaCl in 200g water. Calculate van’t – Hoff factor. The molar depression constant in 1.86 K Kg mol-1
Solution :
Molar mass of solute = \(\frac{1000 \times K_{f} \times \text { mass of } N a C l}{\Delta T_{f} \times \text { mass of solvent }}\)

= \(\frac{1000 \times 1.86 \times 1}{0.24 \times 200}\)
= 38.75 g mol-1

Theoretical molar mass of NaCl is = 58.5 g mol-1

i = \(\frac{\text { Theoretical molar mass }}{\text { Experimental molar mass }}\)
= \(\frac{58.5}{38.75}\) = 1.50

Question 14.
What are the necessary conditions for an ideal solution? Give two examples. For an ideal solution
1. There is no change in volume on mixing two components (solute and solvent)
∆Vmixing = O

2. There is no exchange of heat when the solute is dissolved in solvent (∆Hmixing = 0)

3. Escaping tendency of the solute and the solvent present in it should be the same as in pure liquids.

4. Examples – For ideal solution: Benzene and toluene, n-Hexane and n-Heptane, ethyl bromide and ethyl iodide, chlorobenzene and bromobenzene.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 15.
Explain how non-ideal solutions show positive deviation from Raoult’s law.
Answer:

  1. Let us consider the positive deviation shown by a solution of ethyl alcohol and water.
  2. In this solution, the hydrogen bonding interaction between ethanol and water is weaker than those hydrogen bonding interactions amongst themselves (ethyl alcohol-ethyl alcohol and water-water interaction).
  3. This results in the increased evaporation of both components from the aqueous solution of ethanol.
  4. Consequently, the vapour pressure of the solution is greater than the vapour pressure predicted by Raoult’s law.
  5. Here, the mixing OCCSS is endothermic i.e., (∆Hmixing > O) and there will be a slight increase in volume (∆Vmixing > O)

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-80

Question 16.
Explain with suitable example about negative deviation from law.
Answer:
1. Let us consider a solution of phenol and aniline. Both phenol and aniline form hydrogen-bonding interactions amongst themselves.

2. When mixed with aniline, the phenol molecule forms hydrogen bonding interactions with aniline, which are Stronger than the hydrogen bonds formed amongst themselves.

3. Formation of new hydrogen bonds considerably reduces the escaping tendency of phenol and aniline from the solution.

4. Asa result, the vapour pressure of the solution is less and there is a slight decrease in volume (∆Vmixing < 0) on mixing.

5. During this process evolution of heat takes place i.e., ∆Vmixing < 0 (exothermic).

6. Examples – Acetone + Chloroform, Chloroform + Diethyl ether
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-81

Question 17.
The vapour pressure of a solution containing a nonvolatile, non-electrolyte solute is always lower than that of pure solvent. Give reason.
Answer:
1. The vapour pressure of a solution (P) containing a full volatile solute is lower than that of pure solvent (P°).

2. Consider a closed system is which a pure solvent is in equilibrium with its vapour. At equilibrium the molar Gibbs free energies of solvent in a liquid and gaseous phase are equal (∆G = O).

3. When a solute is added to this solvent the dissolution takes place and its free energy (G) decreases due to increase in entropy.

4. In order to maintain the equilibrium, the free energy of the vapour phase must also decrease.

5. At a given temperature, the only way to lower the free energy of the vapour is to reduce its pressure.

6. Thus the vapour pressure of the solution must decrease to maintain the equilibrium.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 18.
Show that relative lowering of vapour pressure is a colligative property.
Answer:
According to Raoult’s law,
Psolution xA, where xA = mole fraction of solvent.
Psolution = k . xA, where k = proportionality constant
For a pure solvent,
Vapour pressure = P°, xA = 1
solution = k x 1 = k
Substituting P°solvent in Raoult’s law
Psolution = P°solvent . xA
Relative lowenng of vapour pressure
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-82
substituting Psolution as P°xB in the above eaquation
Relative lowering of vapour pressure
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-83
xA + xB = I
where xB = mole fraction of solute. It is clear that the relative lowering of vapour pressure depends only on the mole fraction ofthe solute (xB) and is independent of its nature. Therefore relative lowering of vapour pressure is a colligative property.

Question 19.
Explain why boiling point of solution is greater than that of pure solvent?
Answer:
When a non volatile solute is added to a pure solvent at its boiling point, the vapour pressure of the solution is lowered below 1 atm. To bring the vapour pressure again to I atm the temperature of the solution has to be increased.

As a result, the solution boils at a higher temperature (Tb) than the boiling point of the pure solvent (T°b). This increase in the boiling point is known as elevation of boiling point ∆Tb = Tb – T°b.

Question 20.
Graphically prove that Tb ¡s greater than T°b.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-84
1. The vapour pressure of the solution increases with increase in temperature. The variation of vapour pressure with respect to temperature of pure water is given by the curve – A.

2. At 100°C, the vapour pressure of water is equal to I atm. Hence, the boiling point of water is 100°C (T°b).

3. When a solute is added to water, the vapour pressure of the resultant solution is lowered. The variation of vapour pressure with respect to temperature for the solution is given by curve-B.

4. From the graph, it is evident that the vapour pressure of the solution is equal to 1 atm. pressure at the temperature Tb which is greater than T°b. The difference between these two temperatures (Tb – T°b) gives the elevation of boiling point.
∆Tb = Tbb.

Question 21.
Derive the relationship between the elevation of boiling point and molar mass of non volatile solute.
Answer:
The elevation of boiling point ∆Tb = Tbb.
∆Tb is directly proportional to the concentration of the solute particles.
∆Tb ∝ m, (m = molaLiiy)
∆Tb = kb. m, where kb = ebullioscopic constant
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-85

Question 22.
Define

  1. freezing point
  2. Depression in freezing point.

Explain with graph.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-86
1. Freezing point is defined as the temperature at which the solid and the liquid states of the substances have the same vapour pressure.

2. When a non volatile solute is added to water at its freezing point, the freezing point of water is lowered from 0°C. The lowering of freezing point of the solvent when a solute is added is called depression in freezing point AT1..

3. ∆Tf = T0f – Tf

Question 23.
Define

  1. cryoscopic constant
  2. ebullioscopic constant

Answer:

1. ∆Tf = kf. m, where kf = cryoscopic constant. If m = 1. then ATf = kf
kf is defined as depression in freezing point for 1 molal solution.

2. ∆Tf = kf. m where kf ebullioscopic constant. If m = 1 then ATf = kf
kb is defined as elevation in boiling point for 1 molal solution.

Question 24.
What are the significances of osmotic pressure over other colligative properties ?
Answer:
1. Unlike elevation of boiling point and the depression in freezing point, the magnitude of osmotic pressure is large.

2. The osmotic pressure can be measured at room temperature enables to determine the molecular mass of biomolecules which are unstable at higher temperature.

3. Even for a very dilute solution, the osmotic pressure is large.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 25.
What is haemolysis ? intravenous fluid are isotonic to blood?
Answer:
1. The osmotic pressure of the blood cells is approximately equal to 7 atm at 37°C.

2. The intravenous injections should have saine osmotic pressure as that of the blood (isotonic vith blood).

3. If the intravenous solutions are too dilute that is hypotonie, the solvent from outside of the cells flow into the cell to normalise the osmotic pressure and this process is called haernolysis causes the cells to burst.

4. On the other hand, if the solution is too concentrated, that is hypertonic. the solvent molecules will flow out of the cells,which causes the cells to shrink and die.

5. For this reason, the intravenous fluids are prepared such that they are isotonic to blood (0.9% mass/volume sodium chloride solution).

Question 26.
Explain reverse osmosis.
Answer:
1. The pure water moves through the semipermeable membrane to the NaCl solution due to osmosis.

2. This process can be reversed by applying pressure greater than the osmotic pressure to the solution side. Now the pure water moves from the solution side to the solvent side and this process is called reverse osmosis.

3. Reverse osmosis can be defined as a process in which a solvent passes through a semipermeable membrane in the opposite direction of osmosis, when subjected to a hydrostatic pressure greater than the osmotic pressure.

Question 27.
Explain about the application of reverse osmosis in water purification.
Answer:
1. Reverse osmosis is used in the desalination of sea water and also in the purification of drinking water.

2. When a pressure higher than the osmotic pressure is applied on the solution side (sea water) the water molecules moves from solution side to the solvent side through semi permeable membrane (opposite to osmotic flow). The pure water can be collected.

3. Cellulose acetate (or) polyamide membranes are commonly used in commercial system.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-87

Question 28.
Acetic acid is found to have molar mass as 120 g mol-1. Prove it.
Answer:
1. In certain solvent, solute molecules associate to form a dimer. This reduces the total number of molecules formed in solution and as a result the calculated molar mass will be higher than the actual molar mass.

2. Acetic acid in benzene exist as a dimer
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-88

3. The molar mass of acetic acid calculate using colligative properties is found to be around 120 g mol-1 is two times of the actual molar mass 60 g mol-1.

Question 29.
Depression in freezing point of NaCI is twice that of in urea. Why?
Answer:
1. The electrolyte NaCI dissociates completely into its constituent ions in their aqueous solution. This causes an increase in the total number olparticles present in the solution.

2. When we dissolve 1 mole of NaCI in water it. dissociates and gives 1 mole of Na+ and 1 mole of Cl. Hence the solution will have 2 moles of particles.

But when we dissolve 1 mole of urea (non electrolyte) in water it appears as 1 mole only. So the colligative property value would be double in NaCl than in urea.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 30.
What is van’t Hoff factor? Calculate the van’t Hoff factor value for

  1. acetic acid
  2. NaCl

Answer:
1. van’t Hoff factor is defined as the ratio ofthe actual molar mass to the abnormal (calculated) molar mass of the solute.

2.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-89
van’t Hoff factor (1) for acetic acid = \(\frac { 60 }{ 120 }\) = 0.5

3. van’t Hoff factor (2) for NaCl = \(\frac { 117 }{ 58.5 }\) = 2

Question 31.
Differentiate between ideal solution and non-ideal solution.
Answer:
Ideal solution
An ideal solution is a solution in which each component obeys the Raoult’s law over the entire range of concentration.
For an ideal solution,

  • ∆Hmixing = 0
  • ∆Vmixing = 0

Example: Benzenc and toluene n – Hexane and n – Heptane

Non-ideal solution
The solutions which do not obey Raoult’slaw over the entire range of concentrationsare called non-ideal solution.
For a non-ideal solution.

  • ∆Hmixing \(\quad \neq\) 0
  • ∆Vmixing \(\quad \neq\) 0

Example: Ethyl alcohol and Cyclohexane, Benzene and acetone.

Question 32.
Explain the factors when i = 1, i < 1 and i >1 ?
Answer:
1. For a solute that does not dissociate or associate the vant’s hoff factor is equal to 1 (i = 1) and the molar mass will be close to the actual molar mass.

2. For that solute that associate to form higher oligomers in solution, the van’t Hoff factor will be less than 1 (i < 1) and the observed molar mass will be greater than the actual molar mass.

3. For solutes that dissociates into their constituent ions the van’t Hoff factor will be more than one (i > 1) and the observed molar mass will be less than the normal molar mass.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 33.
State Henry’s law and mention some of its important applications.
Answer:
Henry’s law: The solubility of a gas in a liquid is directly proportional to the pressure of the gas.
Application of Henry’s law:

  1. In the production of carbonated beverages
    (as solubility of CO2 increase at high pressure).
  2. In the deep sea diving.
  3. In the function of lungs.
  4. For climbers or people living at high altitudes,

Question 34.
What type of non – idealities are exhibited by cyclohexane – ethanol and acetone – chloroform mixture? Give reason for your answer.
Answer:
Ideal solutions are those which obey Raoult’s law over extreme range of concentration. Ideal solutions have another important properties:

  • ∆Hmix = 0
  • ∆Vmix = 0

Here-forces of attraction between A – A. B – B and A – B are of the same order. Non ideal solutions do not obey Raoult’s law over the entire range of concentration.
∆Hmixing\(\quad \neq\) 0 and ∆Vmixing\(\quad \neq\) 0

Cyclohexane – ethanol mixture shows positive deviation from Raoult’s law because forces of attraction between cyclohexane and ethanol are less than in between pure cyclohexane as well as pure ethanol.

Acetone-Chloroform mixture shows negative deviation from Raoults law because forces of attraction between acetone and chloroform are higher than that in between pure acetone and pure chloroform molecules.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 35.
Given below is the sketch of a plant for carrying out a process.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-90

  1. Name the process occurring in the above plant.
  2. To which container does the net flow of solvent take place?
  3. Name one SPM which can he used in this plant.
  4. Give one practical use of the plant.

Answer:

  1. Reverse osmosis
  2. In fresh water container from salt water container.
  3. Cellulose acetate is semipermeable membrane (SPM)
  4. Purification of water

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 36.
Define the term osmotic pressure. Describe how the molecular mass of a substance can be determined by a method based on measurement of osmotic pressure?
Answer:
π = CRT
π = \(\frac { n }{ V }\)RT
πV = nRT
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-91
Osmotic pressure is inversely proportional to the molecular mass of the soLute.

Question 37.
1. Menthol is a crystalline substance with peppermint taste. A 6.2% solution of menthol in cyclohexane freezes at – 1.95°C.

Determine the formula mass of menthol. The freezing point and molal depression constant of cyclohexane are 6.5°C and 20.2 K m-1, respectively.

2. State Henry’s Law and mention its two important applications.

3. Which of the following has higher boiling point and why? 0.1 M NaCl or 0.1 M Glucose
Answer:
1.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-92
MB = 158 g mol-1

2. Henry’s Law:
The solubility of a gas in a liquid is directly proportional to the pressure of the gas.
Applications:

  • Solubility of CO2 is increased at high pressure.
  • Mixture of He and O2 are used by deep sea divers because he is less soluble than nitrogen.

3. 0.1 M NaCI, because it dissociates in solution and furnishes greater number of particles per unit volume while glucose being a non-electrolyte does not dissociate.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 38.
Water is a universal solvent. But alcohol also dissolves most of the substances soluble in water and also many more. Boiling point of water is 100°C and that of alcohol is 80°C. The specific heat of water is much higher than the specific heat of alcohol.

  1. List out three possible differences if instead of water as the liquid ¡n our body we had alcohol.
  2. What value can you derive from this special property of water and its innumerable uses in sustaining life on earth?

Answer:
1.
(i) Even a small rise in temperature in the surroundings will raise the temperature of’ the body because the specific heat of alcoholis much less than the specific heat of water. So, in order to cool the body, more sweating will take place.

(ii) As there is less H bonding in alcohol, it will gel evaporated faster. The alcohol will be evaporated at such a faster rate that the liquid has to be ingested all the time.

(iii) Ice which floats on water helps aquatic life to exist even in winter as water insulates the heat from liquid below it to go back to the surroundings. Solid alcohol does not have such special properties.

2. Praise is to the almighty that has so thoughtfully given such special properties to water and made it a liquid that could sustain life on earth.

Question 39.
State Henry’s law for solubility of a gas in a liquid. Explain the significance of Henry’s law constant (KH). At the same temperature, hydrogen is more soluble in water than helium. Which of theni will have a higher value of KH and Why?
Answer:
Henry’s law states that the solubility of a gas in liquid at a given temperature is directly proportional to the partial pressure of the gas.
P = KH x
where P is the pressure of the gas, x is the mole fraction of the gas in the solution and KH is the Henry’s law constant. KH is a function of the nature oIgas.

Higher the value of KH at a given temperature. lower is the solubility of the gas in the liquid. As helium is less soluble in water, so it has a higher value of KH than hydrogen.

Henry’s Law:
As dissolution of agar in liquid is an exothermic process, therefore, the solubility should decrease with in increase in temperature.

Question 40.
What is meant by positive and negative deviations from Raoult’s law and how is the sign ∆Hmix of related to positive and negative deviations from Raoult’s law?
Answer:
Negative deviations:
In these type of deviations, the partial vapour pressure of each component A and B of solution is higher than the vapour pressure calculated from Raoult’s law. For example -Water and ethanol, chloroform and water.

Positive deviations:
In case of positive deviation A – B interactions are weaker than those between A – A or B – B. This means that in such solutions molecules or A (or B) will find it easier to positive deviation from Raoult’s law.

Samacheer Kalvi 11th Chemistry Solutions 5 Marks Questions and Answers

II. Answer the following questions in detail:

Question 1.

  1. Define solution.
  2. Explain the types of solutions with suitable example.

Answer:
1. A solution is a homogeneous mixture of two or more substances, consisting of atoms. ions or molecules.

2. Types and examples of solution.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-93

Question 2.

  1. Define Solubility
  2. Explain the factors that influences the solubility

Answer:

1. The solubility of a substance at a given temperature is defined as the amount of the solute that can be dissolved in 100 g of the solvent at a given temperature to form a saturated solution.

2. Factors influencing solubility
(a) Nature of solute and solvent: Sodium chloride, an ionic compound readily dissolves in polar solvent such as water but it does not dissolve in non polar solvent such as benzene. Most of the organic compounds dissolve in organic solvent and do not dissolve in water.

(b) Effect of temperature: Generally, the solubility of a solid solute in a liquid solvent increases with increase in temperature. The dissolution of NaCl does not vary as the maximum solubility is achieved at normal temperature.

The dissolution of ammonium nitrate is endothermic, the solubility increases with increase in temperature. The dissolution of eerie sulphate is exothermic and the solubility decreases with increase of temperature. In the case of gaseous solute in liquid solvent the soluhility decreases with increase in temperature.

Effect of pressure:
Generally the change in pressure does not have any significant effect in the solubility of solids and l?quids as they are not compressible. However, the solubility of gases generally increases with increase of pressure.

Question 3.
Explain about the factors that are responsible for deviation from Raoult’s law.
Answer:
1. Solute-solvent interactions:
For an ideal solution, the interaction between the solvent molecules (A – A), the solute molecules (B – B) and between the solvent and solute molecules (A – B) are expected to be similar. if these interactions are dissimilar, there will be a deviation from ideal behaviour.

2. Dissolution of solute:
When a solute present in a solution dissociates to give its constituent ions, the resultant ions interact strongly with the solvent and causes deviation from Raoult’s law. e.g., KCI in water deviates from ideal behaviour due to dissociation as K+ and Cl ion which form strong ion-dipole interaction with water molecules.

3. Association of solute:
Association of solute molecules can also cause deviation from ideal behaviour. For example in solution acetic acid exists as a dimer by forming intermolecular hydrogen bonds and hence deviates from Raoult’s law.

4. Temperature:
An increase in temperature of the solution increases the average kinetic energy of the molecules present in the solution which cause decrease in the attractive force between them. As result, the solution deviates from Raoult’s law.

5. Pressure:
At high pressure, the molecules tends to stay close to each other and therefore there will be an increase in their intermolecular attraction. Thus a solution deviates from Raoult’s law at high pressure.

6. Concentration:
When the concentration is increased by adding solute, the solvent-solute interaction becomes significant. This causes deviation from Raoult’s law.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 4.
How would you determine molar mass from relative lowering of vapour pressure.
Answer:
1. The measurement of relative lowering of vapour pressure can be used to determine the molar mass of a non-volatile solute.

2. A known mass of the solute is dissolved in a known quantity of solvent. The relative lowering of vapour pressure is measured experimentally.

3. According to Raoult’s law, the relative lowering of vapour pressure is
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-94
WA = weight of solvent
WB = weight of solute
MA = Molar mass of solvent
MB = molar mass of solute
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-95
where nA = number of moles of solvent
nB = number of moles of solute.
For dilute solution, nA > > nB, nA + nB = nA
Then, Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-96
Number of moles of solvent and solute arc
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-97
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-98
From the above equation, molar mass of the solute MB can be calculated using the known values of WA, WB, MA and the measured relative lowering of vapour pressure.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 5.
How would you determine the molar mass from osmotic pressure.
Answer:
According to van’t Hoff equation
π = CRT
C = \(\frac { n }{ V }\)
Here n = number of moles of solute dissolved in ‘V’ litre of the solution.
π = \(\frac { n }{ V }\) . RT = πV = nRT
If the solution is prepared by dissolving WB of the non-volatile solute in WA g of solvent, then the number of moles of ‘n’ is
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-99
where
MB = molar mass of the solute
Substituting n value, we get
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-100
From the above equation, we can calculate the molar mass of the solute.

Question 6.
What are ideal and non-ideal solutions? Explain with suitable diagram the behaviour of ideal solutions.
Answer:
Ideal solutions:
The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. Ideal solutions are formed by mixing the two components which are identical in molecular size, in structure and have almost identical intermolecular forces.

Examples:

  1. Benzene and toluene
  2. n – Hexane and n-Heptane
  3. Chiorobenzene and bromobenzene.

Characteristics:

  1. They must obey Raoult’s law.
  2. ∆H mixing should be zero.
  3. ∆W mixing should be zero, i.e. volume change on mixing is zero.

Non – ideal solutions:
The solutions which do not obey Raoult’s law are called non-ideal solutions. In case of non – ideal solutions there is a change in volume and heat energy when the two components are mixed.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-101

Characteristics:

  1. They does not obey Raoult’s law.
  2. ∆Vmix \(\quad \neq\) 0
  3. ∆Hmix \(\quad \neq\) 0

Behaviour of Ideal Solutions:
A plot of P1 or P2 versus the mole fraction x1 and x2 for an ideal solution gives a linear plot. These Lines (I and II)pass through the points and respectively whenx1 and x2 is equal to unity.

Similarly the plot (Line III) of Ptotal versus x2 is also linear. The minimum value of is P1° and the maximum value is P2°, assuming that component 1 is less volatile than component 2, i.e. P1° < P20.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 7.
Explain with a suitable diagram and appropriate example, why some non-ideal solution shows positive deviation from Raou It’s law.
Answer:
Some non-ideal solutions show positive deviation from Raoult’s law. Consider a solution of two components A and B.  If A-B interactions in the solution are weaker than the A – A and B – B interactions in the two liquids forming the solution, then the escaping tendency of molecules A and B from the solution become more than in pure liquids.

The total vapour pressure will be greater than the corresponding vapour pressure as expected on the basis of Raoult’s law. This type of behaviour of solution is called positive deviation from Raoult’s law. The boiling point of such solutions are lowered. Mathematically,
PA < PA0 xA
PB < PB0 xB
The total vapour pressure is less than PA + PB
P < PA + PB
P < P°A x xA + P°B xB
Hence
P1 = PA
P2 = PB
Examples of solutions showing positive deviations

  1. Ethyl alcohol and water
  2. Benzene and acetone
  3. Ethyl alcohol and cyclohcxanc
  4. Carbon tetrachloride and chloroform.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-102

Question 8.

  1. What is Osmotic pressure and how is it related to the molecular mass of a non volatile substances?
  2. What advantage the osmotic pressure method has over the elevation of boiling point method for determining the molecular mass?

Answer:
1. Osmotic pressure:
It is the pressure of the solution column that can prevent the entry of solvent molecules through a semi-permeable membrane, when the solution and the solvent are separated by the same. It is denoted by π. Its unit is mm 11g or atmosphere.
We know that, π = CRT
where π is the osmotic pressure and R is the gas constant
π = \(\frac { { n }_{ 2 } }{ V }\) RT
where V is volume of solution per litre containing n2 moles of solute.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-103
By the above relation molar mass of solute can be calculated.

2. The osmotic pressure method has the advantage over other methods as pressure measurement is around the room temperature and molarity of the solution is used instead of molality.

The technique of osmotic pressure for determination of molar mass of solutes is particularly useful for biomolecules as they are generally not stable at higher temperature and polymers have poor solubility.

III. Numerical Problems

Question 1.
Calculate the mole fraction of benzene ¡n solution containing 30% by mass in carbon tetrachioride.
Solution:
30% of benzene in carbon tetrachloride by mass means that Mass of benzene in the solution = 30g
Mass of solution = 100g
Mass of carbon tetrachloride = 100g – 30g = 70g
Molar mass of benzene (C6H6) = 78 g mol-1
Molar mass of CCl4 = 12 + (4 x 35.5) 154g mol-1
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-104

Question 2.
Calculate (he molarity of each of the following solutions:
Solution:

  1. 30 g of CO(NO3)2. 6H2O = in 4.3 L of solution
  2. 30 mL of 0.5 M H2SO4 diluted to 500 mL.

1. Molar mass of CO(NO3)2. 6H2O = 58.7 + 2(14 + 48) + (6 x 8)g mol-1
= 58.7 + 124 + 108g mol-1 = 290.7 gmol-1
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-105
Volume of solution = 4.3 L
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-106

2. 1000 mL of 0.5 M H2SO4 Contain H2SO4 = 0.5 moles
30 mL of 0.5 M H2SO4 contain H2SO4 = \(\frac { 0.5 }{ 1000 }\) x 30 mole = 0.015 mole
Volume of solution = 500 mL = 0.500 L
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-107

Question 3.
Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution.
Solution:
0.25 molal aqueous solution means that
Moles of urea = 0.25 mole
Mass of solvent (water) = 1 kg = 1000 g
Molar mass of urea = 14 + 2 + 12 + 16 + 14 + 2 = 6Og mol-1
0.25 mole of urea = 60 x 0.25 mole = 15 g
Total mass of the solution = 1000 + 15 g = 1015 g = 1.015 g
Thus, 1.015 kg of solution contain urea = 15 g
2.5 kg of solution will require urea = \(\frac { 15 }{ 1.015 }\) x 2.5 kg = 37g

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 4.
H2S, a toxic gas with rotten egg like smell is used for the qualitative analysis. 1f the solubility of H2S in water at STP is 0.195 m. Calculate Henrvs law constant.
Solution:
Solubility of H2S gas = 0.195 m
= 0.195 mole in 1 kg of the Solvent (water)
1 kg of the solvent (water) = 1000 g Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-108
Mole fraction of H2S gas in the solution (x) = \(\frac { 0.195 }{ 0.195 + 55.55 }\) = 0.0035
Pressure at STP = 0.98 7 bar
Applying Henry’s law
PH2S = KH x xH2S
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-109

Question 5.
Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
Solution:
Here
1 = 23.8 mm
W2 = 50g
M2 (urea) = 60g mol-1
W1 = 850g
M1(Water) = 18g mol-1
Here we have to calculate Ps
Applying Raoult’s law,
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-110
Thus, relative lowering of vapour pressure = 0.017
Substituting P° = 23.8 mm Hg
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-111
We get,
23.8 – Ps = 0.017 Ps
Ps = 23.4 mm Hg
Thus, vapour pressure of water in the solution = 23.4 mm Hg

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 6.
Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500g of water such that it boils at 100°C? .
Solution:
Elevation in boiling point required, ∆Tb = 100 – 99.63° = 0.37°
Mass of solvent (water) W1 = 500g
Mass of solute, C12H22O11 = 342 g mol-1
Molar mass of solvent M1 = 18 g mol-1
Applying the formula,
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-112

Question 7.
Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504g mL-1?
Solution:
68% nitric acid by mass means that
Mass of nitric acid = 68 g
Mass of solution = 100 g
Molar mass of HNO3 = 63g mol-1
68g HNO3 = mole = 1.079 mole
Density of solution = 1.504 g mL-1
Volume of solution = \(\frac { 100 }{ 1.504 }\) mL = 66.5 mL = 0.0665 L
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-113

Question 8.
A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate tile mass percentage of the resulting solution.
Solution:
300 g of 25% solution contains solute = 75g
400g of 40% solution contains solute = 160 g
Total mass of solute = 160 + 75 = 235 g
Total mass oÍ’ solution = 300 + 400 = 700 g
% of solute in the final solution = \(\frac { 235 }{ 700 }\) x 1oo = 33.5%
% of water in the finaI solution = 100 – 33.5 = 66.5%

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 9.
A sample of drinking water was found to be severely contaminated with chloroform (CHCI3) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass)

  1. express this in percentage by mass
  2. determine the molality of chloroform in the water sample.

Solution:
1. 15 ppm means 15 parts in million (106) parts by mass in the solution
% of mass = \(\frac { 15 }{ { 10 }^{ 6 } }\) x 100 = 1.5 x 10-4

2. Taking 15g chloroform in 106g of the solution
Mass of the solvent = 106 g
Molar mass of CHCl3 = 12 + 1 + (3 x 35.5) = 119.5 g mol-1
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-114

Question 10.
An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at normal boiling point of the solvent. What is the molar mass of the solute?
Solution:
Vapour pressure of pure water at the boiling point
(P°) = 1 atm 1.013 bar
Vapour pressure of solution Ps = 1.004 bar
M1 = 18 g mol-1
M2 = ?
Mass of solute = W2 = 2g
Mass of solution = 100g
Mass of solvent W1 = 98 g
Applying Raoult’s law for dilute solution
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-115

Question 11.
A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15K.
Solution:
Molar mass of cane sugar
C12H22O11 = 342 g mol-1
Molality of sugar = \(\frac { 5 x 1000 }{ 342 x 100 }\) = 0.146
∆T2 for sugar solution = 273.15 – 271 = 2.15°
∆Tf = Kf x m
Kf = \(\frac { 2.15 }{ 0.146 }\)
Molality of glucose solution = \(\frac { 5 }{ 180 }\) x \(\frac { 1000 }{ 100 }\) = 0.278 m
∆Tf (Glucose) = \(\frac { 2.15 }{ 0.146 }\) x 0.278 = 4.09°K
Freezing point of glucose solution = 273.15 – 4.09 = 269.06 K

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Question 12.
Calculate the amount of benzoic acid (C6HCOOH) required for preparing 250 mL of 0.15 M solution in methanol.
Solution:
0.15 M solution means that 0.15 moles of C6H5COOH is present in IL
= 1000 mL of the solution
Molar mass of C6H5COOH = 72 + 5 + 12 + 32 + 1 = 122 g mol-1
Thus, 1000 mL of solution contains benzoic acid = 18.3g
250 mL of solution will contain benzoic acid
= \(\frac { 18.3 }{ 1000 }\) x 250 = 4.575 g

Question 13.
A solution containing 8 g of a substance in 100 g of diethyl ether boils at 36.86°C, whereas pure ether boils at 35.60 °C. Determine the molecular mass of the solute (For ether Kb 2.02 K kg mol-1)
Solution:
We have, mass of solute, W2 = 8 g
Mass of solvent, W1 = 100 g
Elevation of boiling point
∆Tb = 36.86 – 35.60 = 1.260C
Kb = 2.02
Molecular mass of the solute
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-116
= 128.25g mol-1

Question 14.
Ethylene glycol (molar mass = 62 g mol-1) is a common automobile antifreeze. Calculate the freezing point of a solution containing 12.4 of this substance in 100 g of water. Would it be advisable to keep this substance ¡n the car radiator during summer? (Kf for water = 1.86 k kg/mol-1) (Kb for water = 0.512 K kg/mol-1)
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-117
Since water boils at 100°C, so a solution containing ethylene glycol will boil at 101.024 °C, SO it is advisable to keep this substance in car radiator during summer.

Question 15.
15.0 g of an unknown molecular material is dissolved in 450g of water. The resulting solution freezes at – 0.34°C. What is the molar mass of the material? Kf for water = 1.86 K kg mol-1.
Solution:
W(solute) = 1 5.0 g
W(solvent) = 450 g
∆Tf = T°f – Tf = 0 – ( – 0.34) = 0.34 °C
∆Tf = kf m
0.34 = 1.86 x \(\frac { 15 }{ M }\) x \(\frac { 1000 }{ 450 }\)
M = \(\frac { 1.86 x 15 x 1000 }{ 0.34 x 450 }\) = 182.35 g mol-1

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Common Errors
Common Errors:

  1. Students are writing solute, solvent, students get confused to write A (or) B, 1 (or)2.
  2. Mole, mole fraction may be confused.
  3. In writing osmosis definition Students get confused in mentioning concentration terms.
  4. van’t Hoff factor I formula may be con fused.
  5. Students may get contused rhen they write solute and solvent.
  6. Mole and mole fraction may be confused by students.
  7. Standard solutions must be known.
  8. When students write Raoult’s law, they get confused with solute and solvent.
  9. When they write the definition of osmosis, the conc. term may be little con fused.
  10. van’t Hoff equation may be written wrongly.

Rectifications:

  1. Always solvent is first so it is denoted as A (or) 1 solute is second so, it is denoted as B (or)2.
  2. Mole = n ; Mole fraction x
  3. Osmosis-movenient of solvent from low concentration to high concentration through a semipermeable membrane.
  4. Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-118
  5. For e.g.. solid in liquid means solid is the solute and liquid is the solvent.
  6. Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-119
  7. 1 N = 1 normal solution, 0.1 N Decinormal solution, 0.01 N = Centinormal solution
  8. In Raoult’s law, “A” is always solvent and “B” is always solute.
  9. In osmosis, always solvent rnoes through semi permeable membrane from low concentration to high concentration.
  10. van’t Hoff factory = i
    Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-120

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Students can Download Chemistry Chapter 1 Basic Concepts of Chemistry and Chemical Calculations Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Chemistry Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Samacheer Kalvi 11th Chemistry Chapter 1 Basic Concepts of Chemistry and Chemical Calculations Textual Evaluation Solved

I. Choose the Best Answer

Question 1.
40 ml of methane is completely burnt using 80 ml of oxygen at room temperature. The volume of gas left after cooling to room temperature is ………..
(a) 40 ml CO2 gas
(b) 40 ml CO2 gas and 80 ml H2O gas
(c) 60 ml CO2 gas and 60 ml H2O gas
(d) 120 ml CO2 gas
Answer:
(a) 40 ml CO2 gas
Solution:
CH4(g)  + 2O2(g)  CO2(g) + 2 H2O(l)

Content CH4 O2 CO2
Stoichiometric coefficient 1 2 1
Volume of reactants allowed to react 40 mL 80 mL          –
Volume of reactant reacted and product formed 40 mL 80 mL 40 mL
Volume of gas after cooling to the room temperature

Since the product was cooled to room temperature, water exists mostly as liquid. Hence, option (a) is correct

Question 2.
An element X has the following isotopic composition 200X = 90 %, 199X = 8 % and 202X = 2 %. The weighted average atomic mass of the element X is closest to …………
(a) 201 u
(b) 202 u
(c) 199 u
(d) 200 u
Answer:
(d) 200 u
= \(\frac {(200 × 90) + (199 × 8) + (202 × 2) }{100}\) = 199.96 = 200 u

Question 3.
Assertion:
Two moles of glucose contain 12.044 × 1023 molecules of glucose.

Reason:
The total number of entities present in one mole of any substance is equal to 6.02 × 1022
(a) both assertion and reason are true and the reason is the correct explanation of the assertion
(b) both assertion and reason are true but the reason is not the correct explanation of the assertion
(c) the assertion is true but the reason is false
(d) both assertion and reason are false
Answer:
(c) the assertion is true but the reason is false

Correct reason:
The total number of entities present in one mole of any substance is equal to 6.022 x 1023

Question 4.
Carbon forms two oxides, namely carbon monoxide and carbon dioxide. The equivalent mass of which element remains constant?
(a) Carbon
(b) Oxygen
(c) Both carbon and oxygen
(d) Neither carbon nor oxygen
Answer:
(b) Oxygen

Reaction 1:
2 C + O2 → 2 CO2
2 × 12 g carbon combines with 32 g of oxygen.
Hence, Equivalent mass of carbon = \(\frac {2 × 12}{32}\) × 8 = 6

Reaction 2:
C + O2 → 2 CO2
12 g carbon combines with 32 g of oxygen.
Hence, Equivalent mass of carbon = \(\frac {12}{32}\) × 8 = 3

Question 5.
The equivalent mass of a trivalent metal element is 9 g eq-1 the molar mass of its anhydrous oxide is ………..
(a) 102 g
(b) 27 g
(c) 270 g
(d) 78 g
Answer:
(a) 102 g
Let the trivalent metal be M3+
Equivalent mass = mass of the metal / valance factor
9g eq-1 = mass of the metal / 3 eq
Mass of the metal = 27 g
Oxide formed M2O3
Mass of the oxide = (2 × 27) + (3 × 16) = 102 g

Question 6.
The number of water molecules in a drop of water weighing 0.018 g is ………….
(a) 6.022 × 1026
(b) 6.022 × 1023
(c) 6.022 × 1020
(d) 99 × 1022
Answer:
(c) 6.022 × 1020
Weight of the water drop = 0.0 18 g
No. of moles of water in the drop = Mass of water / molar mass = 0.018/18 = 10-3 mole
No of water molecules present in I mole of water = 6.022 × 1023
No. water molecules in one drop of water (10 moles) = 6.022 × 1023 × 10-3 = 6.022 × 1020

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 7.
1 g of an impure sample of magnesium carbonate (containing no thermally decomposable impurities) on complete thermal decomposition gave 0.44 g of carbon dioxide gas. The percentage of impurity in the sample is ………..
(a) 0 %
(b) 4.4 %
(c) 16 %
(d) 8.4 %
Answer:
(c) 16%
Mg CO3 → MgO + CO2
Mg CO3 : (1 × 24) + (1 × 12) + (3 × 16) = 84 g
CO2 : (1 × 12) + (2 × 16) 44g
100% pure 84 g MgCO3 on heating gives 44 g CO2
Given that I g of MgCO3 on heating gives 0.44 g CO2
Therefore, 84 g MgCO3 sample on heating gives 36.96 g CO2 = 100%
Percentage of purity of the sample = \(\frac{100 \%}{44 \mathrm{g} \mathrm{CO}_{2}}\) × 36.96 g CO2 = 84%
Percentage of impurity = 16%

Question 8.
When 6.3 g of sodium bicarbonate is added to 30 g of the acetic acid solution, the residual solution is found to weigh 33 g. The number of moles of carbon dioxide released in the reaction is-
(a) 3
(b) 0.75
(c) 0.075
(a) 0.3
Answer:
(c) 0.075
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
The amount of CO2 released, x = 3.3 g
No. of moles of CO2 released = 3.3 / 44 = 0.075 mol

Question 9.
When 22.4 liters of H2 (g) is mixed with 11.2 liters of Cl2 (g), each at 273 K at 1 atm the moles of HCl (g), formed is equal to ………..
(a) 2 moles of HCl (g)
(b) 0.5 moles of HCl (g)
(c) 1.5 moles of HCl (g)
(d) 1 moles of HCl (g)
Answer:
(d) 1 moles of HCl (g)
Solution:
H2(g) + Cl2(g) → 2 HCl (g)

Content H2(g) cl2(g) HCl (g)
Stoichiometric coefficient 1 1 2
No. of moles of reactants allowed to react at 273 K and 1 atm pressure 22.4 L (1 mol) 11.2 L (0.5 mol)
No. of moles of a reactant reacted and product formed 0.5 0.5 1
Amount of HCl formed 1 mol

Question 10.
Hot concentrated sulfuric acid is a moderately strong oxidizing agent. Which of the following reactions does not show oxidizing behavior?
(a) Cu + 2H2SO4 → CuSO4 + SO2 + 2H2O
(b) C + 2H2SO4 → 4 CO2 + 2SO2 + 2H2O
(c) BaCl2 + H2SO4 → BaSO4+ 2HCl
(d) none of the above
Answer:
(c) BaCl2 + H2SO4 → BaSO4+ 2HCl
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 11.
Choose the disproportional reaction among the following redox reactions.
(a) 3Mg (s) + N2(g) → Mg2N2 (s)
(b) P4 (s) + 3NaOH + 3H2O → PH3(g) + 3NaH2PO2 (aq)
(c) Cl2 (g) + 2Kl (aq) → 2KC1 (aq) + I2
(d) Cr2O3 (s) + 2Al (s) → A2O3 (s) + 2Cr (s)
Answer:
(b) P4 (s) + 3NaOH + 3H2O → PH3(g) + 3NaH2PO2 (aq)
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 12.
The equivalent mass of potassium permanganate in alkaline medium is
MnO4 + 2H2O + 3e → MnO2 + 4OH
(a) 31.6
(b) 52.7
(c) 79
(d) None of these
Answer:
(b) 52.7
The reduction reaction of the oxidizing agent(MnO4) involves gain of 3 electrons.
Hence the equivalent mass = (Molar mass of KMnO4) / 3 = 158.1 / 3 = 52.7

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 13.
Which one of the following represents 180 g of water?
(a) 5 Moles of water
(b) 90 moles of water
(c) \(\frac{6.022 \times 10^{23}}{180}\) Molecules of water
(d) 6.022 × 1024 Molecules of water
Answer:
(d) 6.022 x 1024 Molecules of water
No. of moles of water present in 180 g
= Mass of water / Molar mass of water
= 180 g /18 g mol-1 = 10 moles
One mole of water contains
= 6.022 × 1023 water molecules
10 mole of water contains = 6.022 × 1023 × 10
= 6.022 × 1024 water molecules

Question 14.
7.5 g of a gas occupies a volume of 5.6 liters at 0°C and 1 atm pressure. The gas is …………
(a) NO
(b) N2O
(c) CO
(d) CO2
Answer:
(a) NO
7.5 g of gas occupies a volume of 5.6 liters at 273 K and 1 atm pressure Therefore, the mass of gas that occupies a volume of 22.4 liters –
\(\frac {7.5 g}{5.6 L}\) × 22. 4 L = 30g
Molar mass of NO (14 + 16) = 30g

Question 15.
The total number of electrons present in 1.7 g of ammonia is ………..
(a) 6.022 × 1023
(b) \(\frac{6.022 \times 10^{22}}{1.7}\)
(c) \(\frac{6.022 \times 10^{24}}{1.7}\)
(d) \(\frac{6.022 \times 10^{23}}{1.7}\)
Answer:
(a) 6.022 × 1023
No. of electrons present in one ammonia (NH3) molecule (7 + 3) = 10
No. of moles of ammonia = \(\frac {Mass}{Molar mass}\)
= \(\frac{1.7 \mathrm{g}}{17 \mathrm{g} \mathrm{mol}^{-1}}\) = 0.1 mol
No. of molecules present in 0ne ammonia
= 0.1 × 6.022 × 1023 = 6.O22 × 1022
No. of electrons present in 0.1 mol of ammonia
10× 6.022 × 1022 = 6.022 × 1023

Question 16.
The correct increasing order of the oxidation state of sulphur in the anions SO42-, SO32-, S2O42-,S2O62- is ………..
(a) SO32- < SO32- < S2O42- < S2O62-
(b) SO42- < S2O42- < S2O62-<SO32-
(c) S2O42- < SO32- < S2O62- < SO42-
(d) S2O62- < S2O42- < SO42- < SO32-
Answer:
(c) S2O42- < SO32- < S2O62- < SO42-
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 17.
The equivalent mass of ferrous oxalate is ……….
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 18.
If Avogadro number were changed from 6.022 × 1023 to 6.022 × 1020, this would change ………..
(a) the ratio of chemical species to each other in a balanced equation
(b) the ratio of elements to each other in a compound
(c) the definition of mass in units of grams
(d) the mass of one mole of carbon
Answer:
(d) the mass of one mole of carbon

Question 19.
Two 22.4 liter containers A and B contains 8 g of O2 and 8 g of SO2 respectively, at 273 K and 1 atm pressure, then ……….
(a) number of molecules in A and B are the same
(b) number of molecules in B is more than that in A
(c) the ratio between the number of molecules in A to the number of molecules in B is 2 : 1
(d) number of molecules in B is three times greater than the number of molecules in A
Answer:
(c) The ratio between the number of molecules in A to number of molecules in B is 2 : 1

Question 20.
What is the mass of precipitate formed when 50 ml of 8.5% solution of Ag NO3 is mixed with 100 ml of 1.865% potassium chloride solution?
(a) 3.59 g
(b) 7 g
(c) 14 g
(d) 28 g
Answer:
(a) 3.59 g
AgNO3 + KCl → KNO3 + AgCl
Solution:
50 mL of 8.5% solution contains 4.25 g of AgNO3
No. of moles of AgNO3 present in 50 mL of 8.5% AgNO3 solution
= Mass / Molar mass = 4.25 / 170 = 0.025 moles
Similarly, No of moles of KCl present in loo mL of 1.865% KCl solution
= 1.865 / 74.5 = 0.025 moles
So total amount of AgCl formed is 0.025 moles (based on the stoichiometry calculator)
Amount of AgCl present in 0.025 moles of AgCl
= no. of moles × molar mass
= 0.025 × 143.5 = 3.59 g

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 21.
The mass of a gas that occupies a volume of 612.5 ml at room temperature and pressure (25°C and 1 atm pressure) is 1.1g. The molar mass of the gas is ………..
(a) 66.25 g mol-1
(b) 44 g mol-1
(c) 24.5 g mol -1
(d) 662.5 g mol-1
Answer:
(b) 44 g mol-1
Solution:
No. of moles of a gas that occupies a volume of 6 12.5 ml at room temperature and pressure
(25° C and 1 atm pressure)
= 612.5 × 10-3 L/24.5 L mol-1
= 0.02 5 moles
We know that,
Molar mass = Mass / no. of moles
= 1.1 g/0.025 mol = 44 g mol-1

Question 22.
Which of the following contain same number of carbon atoms as in 6 g of carbon -12?
(a) 7.5 g ethane
(b) 8 g methane
(c) both (a) and (b)
(d) none of these
Answer:
(c) both (a) and (b)
Solution:
No. of moles of carbon present in 6 g of C – 12 = Mass / Molar mass
= 6/12 = 0.5 moles = 0.5 × 6.022 × 1023 carbon atoms.
No. of moles in 8 g of methane = 8 116 = 0.5 moles
= 0.5 × 6.022 × 1023 carbon atoms.
No. of moles in 7.5 g of ethane = 7.5 / 16 = 0.25 moles
= 2 × 0.25 × 6.022 × 1023 carbon atoms.

Question 23.
Which of the following compound( s) has/have a percentage of carbon same as that in ethylene (C2H4)?
(a) propene
(b) ethyne
(c) benzene
(d) ethane
Answer:
(a) propene
Solution:
Molar mass of carbon
Percentage of carbon in ethylene(C2H6) = Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
= \(\frac {24}{28}\) × 100 = 85.71%
Percentage of carbon in propene (C3H6) = \(\frac {24}{28}\) × 100 = 85.7 1%

Question 24.
Which of the following is/are true with respect to carbon – 12?
(a) relative atomic mass is 12 u
(b) oxidation number of carbon is +4 in all its compounds.
(c) I mole of carbon -12 contain 6.022 × 1022 carbon atoms.
(d) all of these
Answer:
(a) relative atomic mass is 12 u

Question 25.
Which one of the following is used as a standard for atomic mass?
(a) 6C12
(b) 7C12
(c) 6C13
(d) 6C14
Answer:
(a) 6C12

II. Write brief answer to the following questions

Question 26.
Define relative atomic mass.

On the basis of carbon, the relative atomic mass of an element is defined as the ratio of the mass of one atom of the element to the mass of l/12th mass of one atom of Carbon – 12.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 27.
What do you understand by the term mole?
Answer:
The term ‘mole’ is used to represent 6.022 × 1023 entities (atoms or molecules or ions). One mole is the amount of substance of a system, which contains as many elementary particles as there are atoms in 12 g of carbon -12 isotope. The elementary particles can be molecules, atoms, ions, electrons, or any other specified particles.

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 28.
Define equivalent mass.
Answer:
The equivalent mass of an element is the number of parts of the mass of an element which combines with or displaces 1.008 parts of hydrogen or 8 parts of oxygen or 35.5 parts of chlorine.

Question 29.
What do you understand by the term oxidation number?
Answer:
Oxidation number refers to the number of charges an atom would have in a molecule or an ionic compound, if electrons were transferred completely. The oxidation numbers reflect the number of electrons “transferred”.

Question 30.
Distinguish between oxidation and reduction.
Answer:
The oxidation number is defined as the imaginary charge left on the atom when all other atoms of the compound have been removed in their usual oxidation states that are assigned according to set of rules. A term that is often used interchangeably with oxidation number is oxidation state.

Grams to moles calculator is the best mole weight calculator.

Question 31.
Calculate the molar mass of the following compounds.

  1. urea [CO(NH2)2]
  2. acetone [CH3COCH3]
  3. boric acid [H3BO3]
  4. sulphuric acid [H2SO4]

Answer:
1. urea [CO(NH2)2]
Atomic mass of C =12
Atomic mass of O =16
Atomic mass of 2(N) = 28
Atomic mass of 4(H) = 4
∴ Molar mass of Urea = 60

2. Acetone [CH3COCH3]
Atomic mass of 3(C) = 36
Atomic mass of 1(0) = 16
Atomic mass of 6(H) = 6
∴ Molar mass of Acetone = 58

3. Boric acid [H3BO3]
Atomic mass of B = 10
Atomic mass of 3(H) = 3
Atomic mass of 3(O) = 48
∴ Molar mass of Boric acid = 61

4. Sulphuric acid 2[H2SO4]
Atomic mass of 2(H) = 2
Atomic mass of 1(S) = 32
Atomic mass of 4(O) = 64
∴ Molar mass of Sulphuric acid = 98

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 32.
The density of carbon dioxide is equal to 1.977 kg m-3 at 273 K and 1 atm pressure. Calculate the molar mass of CO2
Answer:
Molecular mass = Density x Molar volume
Molar volume of CO2 = 2.24 x 10-2 m3
Density of CO2 = 1.977 kg m-3
Molecular mass of CO2 = 1.977 x 103 gm-3  x 2.24 x 10-2 m3
= 1.977 × 10-1 × 2.24 = 44 g

Question 33.
Which contains the greatest number of moles of oxygen atoms?

  1. 1 mol of ethanol
  2. 1 mol of formic acid
  3. 1 mol of H2O

Answer:
1. 1 mol of ethanol
C2H5OH (ethanol) – Molar mass = 24 + 6 + 16 = 46
46 g of ethanol contains 1 × 6.023 × 1023 number of oxygen atoms.

2. 1 mol of formic acid.
HCOOH (formic acid) – Molar mass = 2+12 + 32 = 46
46 g of HCOOH contains 2 × 6.023 × 1023 number of oxygen atoms.

3. 1 mol of H2O
H2O (water) – Molar mass = 2 + 16 = 18
18 g of water contains 1 × 6.023 × 1023 number of oxygen atoms.
∴ 1 mole of formic acid contains the greatest number of oxygen atoms.

Question 34.
Calculate the average atomic mass of naturally occurring magnesium using the following data

Isotope Isotopic atomic mass Abundance (%)
Mg24 23.99 78.99
Mg26 24.99 10
Mg25 25.98 11.01

Answer:
Isotopes of Mg.
Atomic mass = Mg24 = 23.99 x \(\frac {783. 99}{100}\) = 18.95
Atomic mass = Mg26 = 24.99 x \(\frac {10}{100}\) = 2.499
Atomic mass = Mg25 = 25.98 x \(\frac {11.01}{100}\) = 2.860
Average Atomic mass = 24.309
Average atomic mass of Mg = 24.309

Question 35.
In a reaction x + y + z2 → xyz2, identify the limiting reagent if any, in the following reaction mixtures.
(a) 200 atoms of x + 200 atoms of y + 50 molecules of z2
(b) 1 mol of x + 1 mol of y + 3 mol of z2
(c) 50 atoms of x + 25 atoms of y + 50 molecules of z2
(d) 2.5 mol of x + 5 mol of y + 5 mol of z2
Answer:
x + y + z2
(a) 200 atoms of x + 200 atoms of y + 50 molecules of z2 According to the reaction, 1 atom of x reacts with one atom of y and one molecule of z to give product. In the case (a) 200 atoms of x, 200 atoms of y react with 50 molecules of z2 (4 part) i.e. 50 molecules of z2 react with 50 atoms of x and 50 atoms of y. Hence z is the limiting reagent.

(b) 1 mol of x + 1 mol of y + 3 mol of z2
According to the equation 1 mole of z2 only react with one mole of x and one mole of y. If 3 moles of z2 are there, z is limiting reagent.

(c) 50 atoms of x + 25 atoms of y + 50 molecules of z2
25 atoms of y react with 25 atoms of x and 25 molecules of z2. So y is the limiting reagent.

(d) 2.5 mol of x + 5 mol of y + 5 mol of z2
2.5 mol of x react with 2.5 mole of y and 2.5 mole of z2. So x is the limiting reagent.

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 36.
Mass of one atom of an element is 6.645 × 10-23 g. How many moles of element are there in 0.320 kg?
Answer:
Mass of one atom of an element = 6.645 × 10-23 g = Atomic mass.
Mass of given element = 0.320 kg
Number of moles = Atomic mass
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
= 48.156 x 10-23
= 4.8156 x 10-24 moles.

Question 37.
What is the difference between molecular mass and molar mass? Calculate the molecular mass and molar mass for carbon monoxide.
Answer:
Molecular mass is defined as the ratio of the mass of a molecule to the unified atomic mass unit.
The relative molecular mass of any compound can be calculated by adding the relative atomic masses of its constituent atoms.
Molar mass is defined as the mass of one mole of a substance.
The molar mass of a compound is equal to the sum of the relative atomic masses of its constituents expressed in g mol-1.
Molecular mass of Carbon monoxide
= (1 × Atomic mass of carbon) + ( 1 × Atomic mass of oxygen)
= (1 × 12) +(1 × 16) = 28 u.
Molar mass of Carbon monoxide
= (1 × Atomic mass of carbon) + ( 1 × Atomic mass of oxygen)
= (1 × 12) + (1 × 16) = 28 g mol-1.

Question 38.
What is the empirical formula of the following?

  1. Fructose (C6H12O6) found in honey
  2. Caffeine (C8H10N4O2) a substance found in tea and coffee.

Answer:
1. Fructose (C6H12O6)
Empirical formula is the simplest formula. So it is divided by 6 and so empirical formula is CH2O.

2. Caffeine (C8H10N4O2)
Simplified formula = \(\frac {molecular formula}{2}\)
Empirical formula = C4H5N2O.

Question 39.
The reaction between aluminium and ferric oxide can generate temperatures up to 3273 K and is used in welding metals. (Atomic mass of AC = 21 u Atomic mass of 0 = 16 u) 2Al + Fe2O2 → Al2O3 + 2Fe; If, in this process, 324 g of aluminium is allowed to react with 1.12 kg of ferric oxide.

  1. Calculate the mass of Al2O3 formed.
  2. How much of the excess reagent is left at the end of the reaction?

Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
1. As per the balanced equation 54 g A1 is required for 112 g of iron and 102 g of Al2O3.
54 g of Al gives 102 g of Al2O3.
∴ 324 g of Al will give \(\frac{102}{54}\) x 324 = 612 g of Al2O3.

2. 54 g of Al requires 160 g of Fe2O3 for welding reaction.
∴ 324 g of Al will require \(\frac {160}{54}\) x 324 = 960 g of Fe2O3.
∴ Excess Fe2O3 – Unreacted Fe2O3 = 1120 – 960 = 160 g
160 g of excess reagent is left at the end of the reaction.

Question 40.
How many moles of ethane is required to produce 44 g of CO2 (g) after combustion.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
∴ 44g of CO2 = I mole of CO2
2 moles of CO2 is produced by 1 mole of ethane.
∴ 1 mole of CO2 will be produced by = ?
∴ To produce 1 mole of CO2, the required mole of ethane is = \(\frac {1}{2}\) x 1 = 0.5 mole of ethane.

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 41.
Hydrogen peroxide is an oxidizing agent. It dioxides ferrous ion to ferric ion and reduced itself to water. Write a balanced equation.
Answer:
H2O2 – Oxidizing agent
Fe2+ + H2O2 → Fe3+ + H2O (Acetic medium)
Ferrous ion is oxidized by H2O2 to Ferric ion.
The balanced equation is Fe2+ → Fe3+ + e x 2
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 42.
Calculate the empirical and molecular formula of a compound containing 76.6% carbon, 6.38 % hydrogen and rest oxygen its vapour density is 47.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Empirical formula = C6H6O
Va-pour density 47
∴ Molecular mass = 2 x vapor density = 2 x 47 = 94
Molecular formula Empirical formula x n
Molecular mass x n
n = \(\frac { Molecular mass }{ Empirical formula mass }\) = \(\frac {94}{94}\) = 1
∴ Molecular formula = C6H6O

Question 43.
A Compound on analysis gave Na = 14.31% S = 9.97% H = 6.22% and O = 69.5% calculate the molecular formula of the compound if all the hydrogen in the compound is present in combination with oxygen as water of crystallization, (molecular mass of the compound is 322).
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
All H combines with 10 oxygen atoms to form as 10H2O.
So the empirical formula is Na2SO4 .10H2 O
Empirical formula mass = (23 x 2) + (32 x 1) + (16 x 4) + (10 x 18)
= 46 + 32 + 64 + 180 = 322
n = \(\frac { Molecular mass }{ Empirical formula mass }\) = \(\frac {322}{322}\) = 1
Molecular formula = Na2SO4. 10H2O

Question 44.
Balance the following equations by oxidation number method

  1. K2Cr2 O7 + KI + H2SO2 → K2SO4 + Cr2(SO4)3 + I2 + H2O
  2. KMnO4 + Na2SO3 → Mn02 + Na2SO4 + KOH
  3. Cu+ HNO3 → Cu(N03)2 + NO2 + H2O
  4. H2C2O4 + KMnO4 + H2SO4 → K2SO4 + MnSO4 + CO2 + H2O

Answer:
1. K2Cr2 O7 + KI + H2SO2 → K2SO2 + Cr2(SO4)3 + I2 + H2O
Step – 1.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Step – 2
K2Cr2 O7 + 6KI + H2SO4 → K2SO4 + Cr2(SO4)3 + 3I2 + H2O
Step – 3
To balance other atoms
K2Cr2 O7 + 6KI + H2SO4 → 4K2SO4 + Cr2(SO4)3 + 3I2 + H2O
Step – 4
K2Cr2 O7 + 6KI + 7H2SO4 → 4K2SO4 + Cr2(SO4)3 + 3I2 + 7H2O

2. KMnO4 + Na2SO3 → MnO2 + Na2SO4 + KOH (Alkaline medium)
Step – 1
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Step – 2
2KMnO4 + 3Na2SO3 → 2MnO2 + 3Na2SO4 + KOH
Step – 3
balancing potassium, KOH is multiplied by 2
2KMnO4 + 3Na2SO3 → 2MnO2 + 3Na2SO4 + KOH
Step – 4
To balance H atom, H20 is added on reactant side.
2KMnO4 + 3Na2SO3 + H2O → 2MnO2 + 3Na2SO4 + KOH

3. Cu + HNO3 → Cu(NO3)2 + NO2 + H2O
Step – 1
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Step – 2
Cu + HNO3 → Cu(NO3)2 + NO2 + H2O
Step – 3
To balance Nitrogen, 2HNO3 is multiplied by 2 and NO2 is multiplied by 2
Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + H2O
Step 4.
To balance oxygen, H2O is multiplied by 2
Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O

4. H2C2O4 + KMnO4 + H2SO4 → K2SO4 + MnSO4 + CO2 + H2O
Step – 1
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Step – 2
5 H2C2O4 + KMnO4 + H2SO4 → K2SO4 + MnSO4 + 10 CO2 + H2O
Step – 3
To balance K, KMnO4 and MnSO4 are multiplied by 2
5 H2C2O4 + 2KMnO4 + H2SO4 → K2SO4 + 2MnSO4 + 10 CO2 + H2O
Step – 4
To balance O and H, H2O and H2SO4 are multiplied by 3 and 6.
5 H2C2O4 + 2KMnO4 + 3H2SO4 → K2SO4 + 2MnSO4 + 10 CO2 + 8H2O

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 45.
Balance the following equations by ion-electron method.

  1. KMnO4 + SnCl2  + HCl → MnCl2  + SnCl4  + H2O + KCl
  2. C2O42- + Cr2 O7 2- → Cr 3+ + CO2  (in acid medium)
  3. Na2S2O3 + I2 → Na2S4O6 + NaI (in acid medium)
  4. Zn + NO3 → Zn2+ + NO

Answer:
1. KMnO4 + SnCl2 + HCl → MnCl3 + SnCl3 +H2O + KCl
Oxidation half reaction: (loss of electrons)
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Reduction half reaction: (gain of electrons)
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Add H2O to balance oxygen atoms.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Add HCl to balance hydrogen atoms
KMnO4 + 5e + 8HCl → MnCl2 + 4H2O ………(4)
To equalize the number of electrons equation (1) x 5 and equation (2) x 2
5SnCl2 → 5SnCl4 + 10e
2KMnO4 + 16HCl + 10e → 2MnCl2 + 4H2O + 2KCl
2KMnO4 + 5SnCl2 + 16HCl → 5SnCl4 + 2MnCl2 + 4H2O + 2KCl

2. C2O42- + Cr2 O72- → Cr 3+ + CO2  (in acid medium)
Oxidation half reaction:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Reduction half reaction:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
To balance oxygen atoms, H2O is added on RHS of equation (2)
Cr2O72- + 6e → 2Cr3+ + 7 H2O ……….(3)
To balance Hydrogen atoms, H+ is added on LHS of equation (1)
C2O42- + 14H+ → 2CO2 + 2e ……..(4)
To equalize the number of electrons gained and lost, multiply the equation (4) x 3.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

3. Na2S2O3 + I2 → Na2S4O6 + NaI (in acid medium)
Oxidation half reaction: (Loss of electron)
Na2S2O3 → Na2S4O6 + 2e2- ……..(1)
Reduction half reaction: (Gain of electron)
I2 + 2e2-→ 2NaI …………(2)
Adding (1) and (2)
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
To balance oxygen,
2Na2S2O3 + I2 → Na2S2O2 + 2NaI

In acidic medium
4. Zn + NO3 → Zn2+ + NO
Half reactions are –
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Samacheer Kalvi 11th Chemistry Basic Concepts of Chemistry and Chemical Calculations In Text Questions – Evaluate Yourself

Question 1.
By applying the knowledge of chemical classification, classify each of the following Into elements, compounds or mixtures.

  1. Sugar
  2. Sea water
  3. Distilled water
  4. Carbon dioxide
  5. Copper wire
  6. Table salt
  7. Silver plate
  8. Naphthalene balls

Answer:

Elements Compounds Mixtures
Copper wire (Cu)
Silverplate (Ag)
Sugar
Distilled water Carbon dioxide Naphthalene balls
Seawater
Table salt

Question 2.
Calculate the molar mass of the following.

  1. Ethanol (C2H5OH)
  2. Potassium permanganate (KMnO4)
  3. Potassium dichromate (K2Cr2O7)
  4. Sucrose (C12H22O11)

Answer:
(1) Ethanol (C2H5OH)
Molar mass = (2 × 12) + (6 × 1) + (1 × 16)
= 24 + 6+16
= 46 g mol-1

(2) Potassium permanganate (KMnO4)
Molar mass = 39 + 55 + (4 × 16)
= 39 + 55 + 64
= 158 g mol-1

(3) Potassium dichromate (K2Cr2O7)
Molar mass = (39 × 2) + (2 × 52) + (7 × 16)
= 78 + 104 + 112
= 294 g mol-1

(4) Sucrose (C12H22O11)
Molar mass = (12 × 12) + (22 × 1) + (11 × 16)
= 144 + 22 + 176
= 342 g mol-1

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 3.
(a) Calculate the number of moles present in 9 g of ethane.
Answer:
Mass of ethane = 9 g
The molar mass of ethane C2H6 = 30 g mol-1
No. of moles = \(\frac {Mass}{Molar mass}\) = \(\frac {9}{30}\) = 0.3 mol.

(b) Calculate the number of molecules of oxygen gas that occupies a volume of 224 ml at 273 K and 3 atm pressure.
Answer:
Molar volume of oxygen = 22400 ml.
22400 ml of oxygen contains 6.023 x 1023 molecules.
224 ml of oxygen contain \(\frac{6.023 \times 10^{23}}{22400}\) x 224
\(\frac{6.023 \times 10^{23}}{100}\) = 6.023 × 1021

Question 4.
(a) 0.456 g of metal gives 0.606 g of its chloride. Calculate the equivalent mass of the metal.
Answer:
Mass of the metal = W1 = 0.456 g.
Mass of the metal chloride = W2 = 0.606 g.
∴Mass of chlorine = W2 – W1 = 0.606 – 0.456 = 0.15 g
0.15 g of chlorine combine with 0.456 g of metal.
∴ 35.46 g of chlorine will combine with \(\frac {0.456}{0.15}\) x 35.46 = 107.79 g eq-1

(b) Calculate the equivalent mass of potassium dichromate. The reduction half – reaction in acid medium is –
Cr2O72- + 14H+ +6e → 2 Cr3+ + 7H2O
Answer:
Cr2O72- + 14H+ +6e → 2 Cr3+ + 7H2O
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Molar mass of K2Cr2O7 = 294.18
∴ Equivalent mass of K2Cr2O7 = \(\frac { 294.18}{6}\) = 49.03

Question 5.
A Compound on analysis gave the following percentage composition C = 54.55%, H = 9.09%, O = 36.36%. Determine the empirical formula of the compound.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Empirical formula = C2H4O

Question 6.
Experimental analysis of a compound containing the elements x,y,z on analysis gave the following data, x = 32 %, y = 24 %, z = 44 %. The relative number of atoms of x, y and z are 2,1 and 0.5, respectively. (Molecular mass of the compound is 400 g) Find out.

  1. The atomic masses of the element x,y,z.
  2. Empirical formula of the compound and
  3. Molecular formula of the compound.

Answer:
Element x = 32% ; y = 24% ; z = 44%
Relative number of atoms x = 2 ; y = 1 ; z = 0.5
Molar mass of the compound = 400 g.

1. Atomic mass of the element.
Relative number of atoms Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
∴  Atomic mass Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Atomic mass x = \(\frac {32}{2}\) = 16
Atomic mass y = \(\frac {24}{1}\) = 24
Atomic mass z = \(\frac {44}{0.5}\) = 88
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

2. Empirical formula of the compound x4 y2 Z1
Molecular mass of the compound = 400
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations = \(\frac {400}{200}\) = 2

3. Molecular formula of the compound = x8 y4 z2

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 7.
The balanced equation for a reaction is given below 2x + 3y → 41 + m When 8 moles of x react with 15 moles of y, then –

  1. Which is the limiting reagent?
  2. Calculate the number of products formed.
  3. Calculate the amount of excess reactant left at the end of the reaction.

Answer:
2 x + 3 y → 41 + m

1. 2x reacts with 3y to give products.
5x reacts with 15y means, y is the excess because 8 moles of x should react within 4 x 3y = 12y moles of y to give products. In this reaction 15y moles are used. Therefore, 3 moles of y is excess and it is the limiting agent.

2. When 8 moles of x react with 12 moles of y, the product formed will be 4 x 41 i.e. 161 and 4m as product.
8x + 12y → 161 + 4m

3.  At the end of the reaction, the excess reactant left in 3 moles of y.

Question 8.
Balance the following equation using the oxidation number method
AS2 S3  + HNO3 + H2O → H3ASO4 + H2SO4 + NO
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
To balance oxygen and sulphur, H2O and H2SO4 are added.
3AS2S3 + 2HNO3 + H2O → 6H3ASO4 + 2NO + H2SO2
3AS2S3 + 28HNO3 + 4H2O → 6H3ASO4 + 28NO + 9H2SO4

Samacheer Kalvi 11th Chemistry Solutions Basic Concepts of Chemistry and Chemical Calculations Textual Calculations based on Stolchlometry Solved

Question 1.
How many moles of hydrogen are required to produce 10 moles of ammonia?
Answer:
The balanced stoichiometric equation fòr the formation of ammonia is
N2(g) + 3H2(g) → 2NH2(g)
4s per the stoichiometric equation, to produce 2 moles of ammonia, 3 moles of hydrogen are required.
∴ to produce 10 moles of ammonia,
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
= 15 moles of hydrogen are required.

Question 2.
Calculate the amount of water produced by the combustion of 32 g of methane.
Answer:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
As per the stoichiometric equation,
Combustion of 1 mole (16 g) CH4 produces 2 moles (2 x 18 = 36 g) of water.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Combustion of 32 g CH4 produces
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 3.
How much volume of carbon dioxide is produced when 50 g of calcium carbonate is heated completely under standard conditions?
Answer:
The balanced chemical equation is,
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
As per the stoichiometric equation,
1 mole (100g) CaCO3 on heating produces 1 mole CO2
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
At STP, 1 mole of CO2 occupies a volume of 22.7 litres
At STP, 50 g of CaCO3 on heating produces,
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
= 11.35 litres of CO2

Question 4.
How much volume of chlorine is required to form 11.2 L of HCl at 273 K and 1 atm pressure ?
Answer:
The balanced equation for the formation of HCl is,
H2 (g) + Cl2 (g) → 2 HCl (g)
As per the stoichiometric equation, under given conditions,
To produce 2 moles of HCl, 1 mole of chlorine gas is required.
To produce 44.8 liters of HCl, 22.4 liters of chlorine gas are required
∴ To produce 11.2 liters of HCl,
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
= 5.6 litres of chlorine are required.

Question 5.
Calculate the percentage composition of the elements present in magnesium carbonate. How many kilograms of CO2 can be obtained by heating 1 kg of 90 % pure magnesium carbonate.
Answer:
The balanced chemical equation is
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Molar mass of MgCO3 is 84 g mol-1.
84 g MgCO3 contain 24 g of Magnesium.
∴ 100 MgCO3 contain
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
= 28.57 g Mg i.e. percentage of magnesium = 28.57 %.
84 g MgCO3 contain 12 g of carbon
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
= 14.29 g of carbon
Percentage of carbon = 14.29 %.
84 g MgCO3 contain 48 g of oxygen
∴ 100 g MgCO3 contains
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
= 57.14 g of oxygen.
∴ Percentage of oxygen = 57.14 %.
As per the stoichiometric equation,
84 g of 100 % pure Mg CO3 on heating gives 44 g of CO2.
∴1000 g of 90 % pure Mg CO3 gives
\(\frac {44 g}{84 g x 100 %}\) x 90 % x 1000 g
= 471.43 g of CO2
= 0. 471 kg of CO2

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 6.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
(1) If the entire quantity of all the reactants is not consumed in the reaction which is the limiting reagent ?
(2) Calculate the quantity of urea formed and un reacted quantity of the excess reagent. The balanced equation is
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Answer:
(1) The entire quantity of ammonia is consumed in the reaction. So ammonia is the limiting reagent. Some quantity of CO2 remains unreacted, so CO2 is the excess reagent.

(2) Quantity of urea formed = number of moles of urea formed x molar mass of urea
= 19 moles x 60 g mol-1
= 1140 g = 1.14 kg
Excess reagent leftover at the end of the reaction is carbon dioxide. Amount of carbon dioxide leftover = number of moles of CO2 left over x molar mass of CO2
= 7 moles x 44 g mol-1 = 308 g.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Samacheer Kalvi 11th Chemistry Solutions Basic Concepts of Chemistry and Chemical Calculations Additional Questions Solved

I. Choose the correct answer from the following

Question 1.
Which one of the following is the standard for atomic mass?
(a) 1H1
(b) 66C12
(c) 6C14
(d) 8O16
Answer:
(b) 66C12
Hint:
Standard element used to determine atomic mass is 6 Cl2

Question 2.
One mole of CO2 contains ………….
(a) 6.023 x 1023 atoms of C
(b) 6.023 x 1023 atoms of O
(c) 18.1 x 1023 molecules of CO2
(d) 3g atoms of CO2
Answer:
(a) 6.023 x 1023 atoms of C
Hint:
One mole of any substance contains an Avogadro number of atoms. In this carbon one mole is present and oxygen two atoms are present. So, 6.023 x 1023 atoms of C is correct.

Question 3.
The largest number of molecules is in
(a) 54 g of nitrogen pentoxide
(b) 28 g of carbon dioxide
(c) 36 g of water
(d) 46 g of ethyl alcohol
Answer:
(c) 36 g of water
Hint:(a) 54 g of N2O5
N2O5 = Molecular mass = 28 + 80 = 108
108 g of N2O5 contains 6.023 x 1023 molecules.
∴ 54 g of N2O5 will contain \(\frac{6.023 \times 10^{23}}{108}\) x 54 = 3.0115 x 1023 molecules.

(b) 28 g of CO2
CO2 = Molecular mass = 12 + 32 = 44
44 g of CO2 contains 6.023 x 1023 molecules.
∴ 28 g of CO2 will contain \(\frac{6.023 \times 10^{23}}{44}\) x 28 = 3.832 x 1023 molecules.

(c) 36 g of H2O
H2O = Molecular mass = 2 + 16 = 18
18 g of H2O contains 6.023 x 1023 molecules.
∴ 36 g of H2O will contain \(\frac{6.023 \times 10^{23}}{18}\) x 36 = 12.046 X 1023molecules.

(d) 46 g of C2H5OH
C2H5OH = Molecular mass = 24 + 6 + 16 = 46
46 g of C2H5OH contains 6.023 x 1023 molecules.
So, among the 4, 36 g of water contain the largest number of molecules as 12.046 x 1023.

Question 4.
The number of moles of H2 in 0.224 liter of hydrogen gas at STP is …………..
(a) 1
(b) 0.1
(c) 0.01
(d) 0.001
Answer:
(c) 0.01
22.4 liter of hydrogen gas at STP contains 1 mole.
∴ 0.224 liter of hydrogen gas at STP will contain \(\frac{1}{22.4}\) x 0.224 = 0.01

Question 5.
10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. The amount of water produced in this reaction will be ………..
(a) 3 mol
(b) 4 mol
(c) 1 mol
(d) 2 mol
Answer:
(b) 4 mol
10 g of H2 + 64 g of O2 → water
2H2 + O2 → 2H2O
4 g of hydrogen reacts with 32 g of oxygen. So 10 g of hydrogen will react with 80 g of oxygen. But we have given the amount of oxygen only 64 g. It means, here oxygen is the limiting agent. Now all the oxygen react with 8 g of hydrogen and form 4 moles of water.

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 6.
6.023 x 1020 molecules of urea are present in 100 ml of its solution. The concentration of the solution is –
(a) 0.02 M
(b) 0.1 M
(c) 0.01 M
(d) 0.001 M
Answer:
(c) 0.01 M
100 ml of solution contain 6.023 x 1020 molecules.
6.023 x 1023 molecules in 1000 ml = 1 M.
∴ 6.023 x 1020 molecules in 1000 ml =  Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
= 1020 x 10-23 = 10-3
10-3 moles present in 100 ml.
∴ In 1000 ml the moles present is = \(\frac{10^{-3}}{100}\) x 1000 = 10-2 moles
Concentration = 0.01 M

Question 7.
Two containers A and 13 of the equal volume contain 6 g of O2 and SO2 at 300 K and 1 atm. Then
(a) No. of molecules in A is less than that in B
(b) No. of molecules in A is more than that in B
(c) No. of molecules in A and B are the same
(d) none of these
Answer:
(b) No, of molecules in A is more than that in B
O2 = Mass = 6 g
Molar mass = 32 g
32 g of O2 contains 6.023 x 1023 molecules.
∴ 6 g of O2 will contain = 6.023 X 1023 x 6 = 1.129 x 1023 molecules.
SO2 = Mass = 6 g
Molar mass = 64 g
64 g of SO2 contains 6.023 x 1023 molecules.
∴ 6 g of SO2 will contain 6.023 x 1023 x 6 = 0.5646 x 1023 molecules.
∴ The number of molecules in A is more than that in B.

Question 8.
The number of molecules in 16 g of methane is ……….
(a) 3.023 x 1023
(b) 6.023 x 1023
(c) 16 / 6.023 x 1023
(d) 6.023 / 3 x 1023
Answer:
(b) 6.023 x 1023
Hint:
Methane: CH4
Molecular mass 12 + 4 = 16
16 g of methane contains Avogadro’s number of molecules 6.023 x 1023 molecules.

Question 9.
The number of atoms in 4.25 g of ammonia is …………..
(a) 1 x 1023
(b) 2 x 1023
(e) 4 x 1023
(d) 6 x 1023
Answer:
(d) 6 x 1023
Hint:
Ammonia = NH3 (4 atoms)
Molecular mass = 14 + 3 = 17
17 g of Ammonia contains 6.023 x 1023 atoms.
∴ 4.25 g of Ammonia will contain 6.0231; 1023 x 4.25 = 1.5055 x 1023 molecules.
∴ 1.5055 x 1023 molecules will contain 4 x 1.5 x 1023 = 6 x 1023 molecules.

Question 10.
The number of molecules in a drop of water (0.0018 ml) at room temperature is ……….
(a) 6.02 x 1023
(b) 1.084 x 1023
(c) 4.84 x 1023
(d) 6.02 x 1023
Answer:
(a) 6.02 x 1023
Hint:
0.0018 ml – drop of water = 0.0018 g
H2O = molecular mass = 18 g.
Number of molecules in 18 g = 6.023 x 1023
∴ Number of molecules in 0.0018 g 6.023 x 1023 x 0.0018 = 6.023 x 1023 x 105
= 6.023 x 1019 molecules.
or
Density of water at 25°C = 997.0479 g / L
Mass of 0.0018 ml (or) 0.00 18 x 10-3 L
= D x V
= 997.05 x 0.0018 x 10-3
= 1.795 x 10-3 g
Molar mass of water = 18 g
Mole = \(\frac{Mass}{Molecular mass}\) = \(\frac{1.795 \times 10^{-3}}{18}\)
= 9.971 x 10-5
∴ Number of molecules in 0.0018 ml = moles x Avogadro number
= 9.971 x 10-5 x 6.023 x 1023
= 6 x 1019 molecules.

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 11.
7.5 g of gas occupies 5.6 liters of volume at STP. The gas ……….
(a) NO
(b) N2O
(c) CO
(d) CO2
Answer:
(a) NO
Hint:
22.4 liters = 1 mole
5.6 liters = \(\frac {1}{22.4}\) x 5.6 = 0.25 mole.
NO = molar mass = 14 + 16 = 30 g = 1 mole
N2O = molar mass = 28 + 16 = 44g = 1 mole
CO = molar mass = 12 + 16 = 28 g = 1 mole
CO2=molar mass = 12 + 32 = 44g = 1 mole
Among the four gases, 0.25 mole = 7.5 g is equal to NO gas.

Question 12.
The mass in grams of 0.45 mole of CO2 ions ……….
(a) 1.8
(b) 40
(c) 36
(d) 18
Answer:
(d) 18
Hint:
Ca = Atomic mass = 40
Ca → Ca2+ + 2e
41 g of Ca = 1 mole (for Ca2+ Atomic mass remains same)
1 mole of Ca2+ = 40 g
∴ 0.45 mole of Ca2+ = \(\frac {40}{1}\) x 0.45 = 18g

Question 13.
The mass of one molecule of HI in grams is ……….
(a) 2.125 x 10-22
(b) 128
(c) 127
(d) 6.02 x 10-23
Answer:
(b) 128
Hint:
HI = 1 mole = 1 + 127 = 128 g

Question 14.
Avogadro’s number is the number of molecules present in ……….
(a) 1 g of molecule
(b) 1 g atom of molecule
(c) gram molecular mass
(d) I lit of molecule
Answer:
(c) gram molecular mass
by definition (c) is correct

Question 15.
Which of the following contains same number of carbon atoms as are in 6.0 g of carbon (C-12)?
(a) 6.0 g ethane
(b) 8.0 g methane
(c) 21.0 g Propane
(d) 28.0 g CO
Answer:
(b) 8.0 g methane

Hint:
(a) 6.0 g of ethane (C2H6)
C2H6 = molar mass = 24 + 6 = 30 g
30 g of ethane contains 2 x 6.023 x 1023 Carbon atoms.

(b) 8.0 g of methane (CH4)
CH4 molar mass = 12 + 4 = 16g
16 g of methane contains 6.023 x 1023 Carbon atoms.

(c) 21.0 g of propane (C3H8)
C3H8 = molar mass = 36 + 8 = 44 g
44 g of propane contains 3 x 6.023 x 1023 Carbon atoms.

(d) 28.0 g of Carbon monoxide (CO)
CO = molar mass = 12+ 16 = 28 g
28 g of Carbon monoxide contains 6.023 x 1023 Carbon atoms.
6.0 g of Carbon contains = 6.023 x 10 x 6 = 3.0115 x 1023 Carbon atoms.
Among the (a), (b), (c), (d) – 8 g of CH4 contains x 8 = 3.0115 x 1023 Carbon atoms.

Question 16.
The equivalent mass of KMnO4 when it is converted to MnSO4 is equal to molar mass divided by ………..
(a) 6
(b) 4
(c) 5
(d) 2
Answer:
(c) 5
Hint:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 17.
How many equivalents of Sodium sulphate are formed when Sulphuric acid is completely neutralized with a base NaOH?
(a) 0.2
(b) 2
(e) 0.1
(d) 1
Answer:
(d) 1
Hint:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 18.
Cl2 changes to Cl and ClO in cold NaOH. Equivalent mass of Cl2 will be ………..
(a) Molar mass / 2
(b) Molar mass / 1
(c) Molar mass / 3
(d) 2 x Molar mass / 2
Answer:
(a) Molar mass / 2
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 19.
Equivalent mass of KMnO4 in acidic medium, concentrated alkaline medium and dilute basic medium respectively are M, M, M. Reduced products can be ……………
(a) MnO2, MnO22-, Mn2+
(b) MnO2, Mn2+, MnO42-
(c) Mn2+, MnO2, MnO42-
(d) Mn2+, MnO42-, MnO2
Answer:
(c) Mn2+, MnO2, MnO42-
Hint:
MnO4 + 8H+ + 5e → Mn2+ + 4H2O (acidic medium)
MnO4 + 4H+ + 3e → MnO2 + 2H2O (concentrated basic medium)
MnO4 + e → MnO42- (dilute basic medium)

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 20.
The empirical formula of hydrogen peroxide is
(a) HO
(b) H2O
(e) H3O
(d) H2O2
Answer:
(a) HO
Hint:
Molecular formula of hydrogen peroxide = H2O2
H2O2 ÷ 2 = HO = Empirical formula.

Question 21.
Molecular mass =
(a) Vapour Density × 2
(b) Vapour Density ÷ 2
(c) Vapour Density × 3
(d) Vapour Density
Answer:
(a) Vapour Density × 2

Question 22.
20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What will be the percentage of purity of magnesium carbonate in the sample?
(a) 60
(b) 84
(e) 75
(d) 96
Answer:
(b) 84
Hint:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
84 g of MgCO3 gives 40 g of MgO. (100% purity)
20 g of MgCO3 will give = \(\frac{40 \times 20}{84}\) = 9.52 g
9.52 g of MgO is given by 100% pure MgCO3.
8.0 g of MgO will be given by = \(\frac{100 \times 8}{9.52}\) = 84.03%

Question 23.
What is the mass of the precipitate formed when the preparation of alkyl halides 50 ml of 16.9 % solution of AgNO3 is mixed with 50 ml of 5.8 % NaCl solution?
(a) 7 g
(b) 14 g
(c) 28 g
(d) 35 g
Answer:
(a) 7 g
Hint:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 24.
When 22 L of hydrogen gas is mixed with 11.2 L of chlorine gas, each at STP, the moles of HCl gas formed is equal to ……….
(a) 2
(b) 0.5
(c) 1.5
(d) 1
Answer:
(d) 1
Hint:
H2 + Cl2 → 2HCl
22 liters + 11.2 litres
1 mole + \(\frac {1}{2}\) mole = 1 mole of HCl and \(\frac {1}{2}\) mole of H2 is remained.

Question 25.
5.6 L of a gas at STP are found to have mass of 11 g. The molecular mass of the gas is
(a) 36
(b) 48
(c) 40
(d) 44
Answer:
(d)44
Hint:
5.6 L of gas has the mass of 11 g.
∴ 22.4 L of gas will have the mass \(\frac {11}{5.6}\) x 22.4 = 44

Question 26.
Oxidation number f Fluorine in all compounds is ……….
(a) +1
(b) -1
(c) 0
(d) -2
Answer:
(b) -1

Question 27.
In redox reaction which of the following is true?
(a) Number of electrons lost is more than number of electrons gained
(b) Number of electrons lost is less than number of electrons gained
(c) Number of electrons lost s equal number of electrons gained
(d) No transfer and gain of electrons during the reaction.
Answer:
(c) Number of electrons lost is equal number of electrons gained

Question 28.
Which of the following is a mono-atomic molecule?
(a) Hydrogen
(b) Oxygen
(c) Sodium
(d) Ozone
Answer:
(c) Sodium

Question 29.
Which one of the following is a diatomic molecule?
(a) Ozone
(b) Copper
(c) Hydrogen
(d) Gold
Answer:
(c) Hydrogen

Question 30.
The value of Avogadro Number N is equal to ……….
(a) 2.24 x 10-2L
(b) 22400 cm3
(c) 6.023 x 10-23
(d) 6.023 x 1023
Answer:
(d) 6.023 x 1023

Question 31.
46 g of ethanol contains ……….
(a) 2 x 6.023 x 1023 C atoms
(b) 3 x 6.023 x 1023 atoms
(c) 9 x 6.023 x 1023 H atoms
(d) 6.023 x 1023 Carbon atoms
Answer:
(a) 2 x 6.023 x 1023 C atoms
Hint:
C2H5OH = Molecular mass = (12 x 2) + (1 x 6) + (1 x 16)
= 24 + 6 + 16
= 46
2 Carbon atoms are present.
∴ 2 x 6.023 x 1023 C atoms is correct.

Question 32.
The mass of one mole of CaCl2 is ……….
(a) 55.5 g mol-1
(b) 111 g mol-1
(c) 222 g mol -1
(d) 77.5 g mol-1
Answer:
(b) 111 g mol-1
Hint:
CaCl2 = Mass = 40 + 71 = 111 g mol-1

Question 33.
22 g of CO2 contains molecules of CO2
(a) 6.023 x 1023
(b) 6.023 x 1023
(c) 3.0115 x 1023
(d) 3.0115 x 1023
Answer:
(c) 3.0115 x 1023
Hint:
44 g of CO2 contains 6.023 x 1023 molecules.
∴ 22 g of CO2 will contain = \(\frac{6.023 \times 10^{23}}{44}\) x 22 = 3.0115 x 1023

Question 34.
The formula weight of ethanol (C2H5OH) is ……….
(a) 56.5 amu
(b) 16 amu
(c) 60 amu
(d) 46 amu
Answer:
(d) 46 amu
Hint:
Formula weight of C2H5OH = (12 x 2) + (6 x 1) + (1 x 16)
= 24 + 6 + 16 = 46 amu

Question 35.
The number of moles of ethane in 60 g is ……….
(a) 2
(b) 4
(c) 0.5
(d) 1
Answer:
(a) 2
Hint:
C2H6 – Ethane – molar mass = 24 + 6 = 30 g
30 g of C2H6 contains = 1 mole.
∴ 60 g of C2H6 will contains = 2 x 1 = 2 moles.

Question 36.
Which of the following method is used to prevent rusting of iron?
(a) Galvanization
(b) Painting
(c) Chrome plating
(d) all the above
Answer:
(d) all the above

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 37.
Which of the following is not a redox reaction?
(a) H2 + F2 → 2HF
(b) Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O
(c) 2H2 + O2 → 2H2O
(d) AgCl + NH3 → [Ag(NH3)2]Cl
Answer:
(d) AgCl + NH3 → [Ag(NH3)2]Cl

Question 38.
How many H2O molecules are there in a snowflake weighing 1 mg?
(a) 3.35 x 1019
(b) 6.023 x 1023
(c) 335 x 10-19
(d) 100
Answer:
(a) 3.35 x 1019
Hint:
1 mg of H2O
Molar mass of H2O = 2 + 16 = 18 g
1 Mole contains 6.023 x 1023 water molecules
18 g contains 6.023 x 1023 water molecules
l mg contains \(\frac{6.023 \times 10^{23}}{18}\) x 1g /1000 mg = 3.35 x 10-19 H2O molecule.

Question 39.
The volume of HCl gas weighing 73 g at STP is ……….
(a) 2.24 x 10-2 m3
(b) 4.48 x 10-2 m3
(c) 4.48 x 102m3
(d) 2.24 x 102m3
Answer:
(b) 4.48 x 10-2 m3
Hint:
HCl = Molar mass = 1 + 35.5 = 36.5 g
36.5 g of HCl occupies 2.24 x 10-2m3
∴ 73 g of HCl at STP will occupy \(\frac{2.24 \times 10^{-2}}{36.5}\) x 73 = 4.48 x 10-2m3

Question 40.
The molar volume of 22 g of CO2 is ……….
(a)2.24 x 10-2m3
(b)4.48 x 10-2m3
(c) 1.12 x 10-2m3
(d)2.24 x 10-2m3
Answer:
(c) 1.12 x 10-2m3
Hint:
CO2 = Molar’mass of 12 + 32 = 44 g
44 g of CO2 occupies molar volume = 2.24 x 10-2m3
∴ 2g of CO2 will occupy = \(\frac{2.24 \times 10^{-2}}{44}\) x 2= 1.12 x 10-2 L

Question 41.
The equivalent mass of Aluminium is ……….
(a) 27
(b) 13.5
(c) 54
(d) 9
Answer:
(d) 9

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 42.
The equivalent mass of HSO4 is ……….
(a) 98
(b) 97
(c) 48
(d) 96
Answer:
(a) 98
Hint:
HSO4 = Molar mass= 1 + 32 + 64 + 1 = 98
Equivalent mass = Molar mass / 1 = 98

Question 43.
The equivalent mass of NaCl is ……….
(a) 40
(b) 58.5
(c) 35.5
(d) 23
Answer:
(b) 58.5
Hint:
NaCl = Salt Molar mass 23 + 35.5 = 58.5
Equivalent mass of Salt = Molar mass of Salt.

Question 44.
Match the List-I and List-lI using the correct code given below the list.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 45.
Match the List-I and List-Il using the correct code given below the list.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 46.
Consider the following statements ……….
(i) Empirical formula shows the actual number of atoms of different elements in one molecule of the compound.
(ii) Ozone is a diatomic molecule.
(iii) Gases are easily compressible.
Which of the above statement is/are not correct?
(a) (i), (ii), (iii)
(b) (i) & (ii)
(c) (ii) & (iii)
(d) (iii) only
Answer:
(b) (i) & (ii)

Question 47.
How many molecules of hydrogen is required to produce 4 moles of ammonia?
(a) 15 moles
(b) 20 moles
(c) 6 moles
(d) 4 moles
Answer:
(c) 6 moles
Hint:
3H2 + N2 → 2NH3
To get 2 moles of ammonia, 3 mole of H2 is required.
To get 4 moles of ammonia = \(\frac {3}{2}\), x 4
= 6 moles of H2 is required.

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 48.
The number of moles of oxygen required to prepare 1 mole of water is …………..
(a) 1 mole
(b) 0.5 mole
(c) 2 moles
(d) 0.4 mole
Answer:
(b) 0.5 mole
Hint:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations - 147
0.5 mole of oxygen is required to prepare 1 mole of H2O.

Question 49.
How much volume of CO2 is produced when 50 g of CaCO3 is heated strongly?
(a) 2.24 x 10-2 m3
(b) 22.4
(c) 11.2 L
(d) 22400 cm3
Answer:
(c) 11.2 L
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations - 148
100 g of CaCO3 produces 22.4 liters of CO2.
50 g of CaCO3 will produce = \(\frac{22.4}{100} \times 50\) = 11.2 litres

Question 50.
Which one of the following is not a redox reaction?
(a) Rusting of iron
(b) Extraction of metal Na
(c) Electroplating
(d) Aluminothermic process
Answer:
(a) Rusting of Iron

Question 51.
In the reaction 2 AuCl3 + 3 SnCl2 → 2 Au + 3 SnCl4 which is an oxidising agent?
(a) AuCl3
(b) Au
(c) SnCl2
(d) Both AuCl3 and SnCl2
Answer:
(a) AuCl3
Hint:
AuCl3 undergoes reduction. So, it is an oxidising agent.

Question 52.
Identify the compound formed during the rusting of iron.
(a) Fe2O3
(b) Fe2O3. x H2O
(c) FeO. x H2O
(d) FeO
Answer:
(b) Fe2O3. x H2O – Hydrated iron oxide is rust.

Question 53.
The oxidation state of a substance in its elementary state is equal to ………..
(a) -1
(b) -2
(c) zero
(d) charge of the ion
Answer:
(c) zero

Question 54.
The oxidation number of fluorine in all its compounds is equal to ………….
(a) -1
(b) +1
(c) – 2
(d) +2
Answer:
(a) -1

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 55.
Consider the following statements.
(i) The sum of the oxidation number of all the atoms in a neutral molecule is equal to zero.
(ii) Fluorine has an oxidation number +1 in all its compounds.
(iii) The oxidation number of a substance in its elementary state is equal to zero.
Which of the above statement is/are not correct?
(a) (i), (ii) & (iii)
(b) (ii) & (iii)
(c) (i) only
(d) (ii) only
Answer:
(d) (ii) only

Question 56.
The oxidation number of Cr in K2Cr2O7 is ………….
(a) +4
(b) + 6
(c)  O
(d) + 7
Answer:
(b) + 6
K2Cr2O2
2 + 2x – 14 = 0
2x – 12 = 0
2x = + 12
x = + 6

Question 57.
The oxidation number of N in NH4 ion is ……….
(a) +4
(b) + 3
(c) – 3
(d) – 4
Answer:
(c) – 3
NH2+
x + 4 = + 1
x = + 1 – 4
x = – 3

Question 58.
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s). In this reaction, which gets oxidised?
(a) Cu2+
(b) Zn2+
(c) Zn
(d) Zn, Cu2+
Answer:
(c) Zn
Hint:
Zn → Zn2+ + 2e (loss of electron = oxidation)
Zn gets oxidised

Question 59.
Which one of the following is an example of disproportionation reaction?
(a) CuSO4 + Zn → ZnSO4 + Cu
(b) 2KClO3 → 2KCI + 3O2
(c) PCl5 → PCl3 + Cl2
(d) 4H3PO3 → 3H3PO4 + PH3
Answer:
(d) 4H3PO3 → 3H3PO4 + PH3 (Auto oxidation and reduction reaction)

Question 60.
The number of molecules in 40 g of sodium hydroxide is ……….
(a) 6.023 x 1023
(b) 3.0115 x 1023
(c) 6.023 x 1023
(d) 2 x 6.023 x 1023
Answer:
(c) 6.023 x 1023
Sodium hydroxide = NaOH = 23 + 16 + 1 = 40 g
40g = 1 mole = 6.023 1023

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 61.
The mass of one molecule of AgCl in grams is ……….
(a) 108 g
(b) 143.5 g
(c) 35.5 g
(d) 243.5 g
Answer:
(b) 143.5 g
Hint:
Mass of AgCl = 108 + 35.5 = 143.5 g.

Question 62.
The empirical formula of Alkene is ……….
(a) CH
(b) CH2
(c) CH3
(d) CH3O
Answer:
(b) CH2
Hint:
Alkene CnH2n Molecular formula
E.F. = M.F./2
∴ Empirical formula = CH2

Question 63.
22 g of a gas occupies 11.2 liters of volume at STP. The gas is ……….
(a) CH4
(b) NO
(c) CO
(d) CO2
Answer:
(d) CO2
Hint:
22 g of a gas occupies 11.2 liters.
11.2 liters is occupied by 22 g of gas.
∴ Molar volume 22.4 liter will be occupied by \(\frac {22}{11.2}\) x 22.4 = 44 g
∴ The gas is CO2.

Question 64.
The number of moles of H2 in 2.24 liter of hydrogen gas at STP is ……….
(a) 1
(b) 0.1
(c) 0.01
(d) 0.001
Answer:
(b) 0.1
Hint:
22.4 liter = 1 molar volume = 1 mole.
∴ 2.24 liter = \(\frac {1}{22.4}\) x 2.24 = 01 mole

Question 65.
How many molecules are present in 32 g of methane?
(a) 2 x 6.023 x 1023
(b) 6.023 x 1023
(c) 6.023 x 1023
(d) 3.011 x 1023
Answer:
(a) 2 x 6.023 x 1023
Hint:
Methane (CH4) – Molar mass = 12 + 4 = 16g.
16 g contains 6.023 x 1023 molecules.
∴ 32 g of methane will contain = \(\frac{6.023 \times 10^{23}}{16} \times 32^{2}\) = 2 x 6.023 x 1023

Question 66.
The empirical formula of glucose is ……….
(a) CH
(b) CH2O
(c) CH2O2
(d) CHO
Answer:
(b) CH2O
Hint:
Glucose = Molecular formula = C6H12O6
Empirical formula = \(\frac { Molecular formula}{6}\) = \(\\frac{\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}{6}\) = CH2O

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 67.
How many moles of water is present in I L of water?
(a) 1
(b) 18
(c) 55.55
(d) 5.555
Answer:
(c) 55.55
Hint:
1 Liter of water = 1000 g.
The molar mass of water = 18 g
Number of moles = \(\frac {Mass}{Molar mass}\) = \(\frac {1000}{18}\) = 55.55 moles

Question 68.
How many moles of hydrogen atoms are present in 1 mole of C2H6?
(a) 18 moles
(b) 6 moles
(c) 3 moles
(d) 1 mole
Answer:
(b) 6 moles
Hint:
C2H6 contains 6H atoms. 6 moles.

Question 69.
The molar mass of Na2SO4 is ……….
(a).129
(b) 142
(c) 110
(d) 70
Answer:
(b) 142
Hint:
Na2 SO4 = Molar mass
= (23 x 2) + (32 x 1) + (16 x 4)
= 46 + 32 + 64
= 142

Question 70.
Match the List-I with List-Il using the correct code given below the list.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 71.
Ore mole of CO2 contains ………….
(a) 6.02 x 1023 atoms of C
(b) 3 g of CO2
(c) 6.02 1023 atoms of O
(d) 18.1 x 1023 molecules of CO2
Answer:
(a) 6.02 x 1023 atoms of C

Question 72.
5.6 liters of oxygen at STP is equivalent to ……….
(a) 1 mole
(b) 1/4 mole
(c) 1/8 mole
(d) 1/2 mole
Answer:
(b) 1/4 mole
Hint:
22.4 litres of O2 = 1 mole
∴ 5.6 litres of O2 = \(\frac {1}{22.4}\) x 5.6 = 0.25 mole = 1/4 mole.

Question 73.
How many grams are contained in 1 gram atom of Na?
(a) 13 g
(b) 1 g
(c) 23 g
(d) 1/23 g
Answer:
(c) 23 g
Hint:
1 gram atom of Na
Na = Atomic mass 23 g (or) 23 amu
1 gram atom of Na = 1 mole = 23 g.

Question 74.
12 g of Mg will react completely with an acid to give ……….
(a) 1 mole of O2
(b) 1/2 mole of H2
(c) 1 mole of H2
(d) 2 mole of H2
Answer:
(b) 1/2 mole of H2
Hint:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
∴ 12 g of Mg \(\frac {1}{2}\) mole of H2O

Question 75.
which of the following has the highest mass?
(a) I g atom of C
(b) 1/2 mole of CH4
(c) 10 ml of water
(d) 3.011 x 1023 atoms of oxygen
Answer:
(a) I g atom of C
(a) 1 g atom of C = 12 g
(b) 1/2 mole of CH4 = \(\frac {12 + 4}{2}\) = 8 g.
(c) 10 ml of water (H2O) = 1 x 10 = 10 g
(d) 3.011 x 1023 atoms of oxygen = 0.5 mole of oxygen = 8 g

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 76.
The empirical formula of sucrose is ……….
(a) CH2O
(b) CHO
(c) C12H22O11
(d) C(H2O)2
Answer:
(a) CH2O
Hint:
Sucrose Molecular formula = C12H22O11
E.F.= \(\frac{\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}}{12}\) = CH2O

Question 77.
The number of grams of oxygen in 0.10 mol of Na2CO3. 10H2O is ………..
(a) 20.8 g
(b) 18 g
(c) 108 g
(d) 13 g
Answer:
(a) 20.8 g
Hint:
Na2CO2.10H2O = 1 mole
1 mole of Na2CO3. 10H2O contains 13 oxygen atoms.
Mass of 13 oxygen atoms = 13 x 16 = 208
1 mole of Na2CO3.10H2O contains 208 g of oxygen.
∴ 0.10 mole of Na2CO3.10H2O contains \(\frac {208}{1}\) x 0.12 = 08 g.

Question 78.
The mass of an atom of nitrogen is ……….
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 79.
Which of the following halogens do not exhibit positive oxidation numbers in their compounds?
(a) Fluorine
(b) Chlorine
(c) Iodine
(d) Bromine
Answer:
(a) Fluorine

Question 80.
Which of the following is the most powerful oxidizing agent?
(a) KMnO4
(b) K2Cr2O7
(c) O3
(d) H2O2
Answer:
(a) KMnO4

Question 81.
On the reaction 2Ag + H2SO4 → Ag2SO4 + 2H2O + SO2. Sulphuric acid acts as ……………
(a) oxidizing agent
(b) reducing agent
(c) a catalyst
(d) an acid as well as an oxidant
Answer:
(d) an acid as well as an oxidant

Question 82.
The oxidation number of carboxylic carbon atom in CH3COOH is ……….
(a) + 2
(b) + 4
(c) + 1
(d) + 3
Answer:
(d) + 3
CH3COOH
-3 + 3 + x – 4 + 1
x – 3 = 0
x = + 3
Carboxylic carbon oxidation number = + 3

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 83.
When methane is burnt in oxygen to produce CO2 and H2O, the oxidation number of carbon changes by ……….
(a) – 8
(b) + 4
(c) zero
(d) + 8
Answer:
(b) + 4
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 84.
The oxidation number of carbon is zero in ……….
(a) HCHO
(b) C12H22O11
(c) C6H12O6
(d) all the above
Answer:
(d) all the above

Question 85.
The oxidation number of Fe in Fe2(SO4)3 is ……….
(a) + 2
(b) + 3
(c) + 2, + 3
(d) O
Answer:
(b) + 3
Fe2(SO4)3
2x – 6 = 0
x = + 3

Question 86.
Among the following molecules in which Chlorine shows maximum oxidation state?
(a) Cl21
(b) KCl
(c) KClO3
(d) Cl2O7
Answer:
(d) Cl2O7
Cl2O7
2x – 14 = 0
2x = + 14
x = + 7

Question 87.
The oxidation number of carbon in CH3 → CH2OH is ………….
(a) + 2
(b) – 2
(e) O
(d) + 4
Answer:
(b) – 2
C2H5OH
2x +6 – 2 = O
2x + 4 = 0
2x = -4
x = -2

Samacheer Kalvi 11th Chemistry Basic Concepts of Chemistry and Chemical Calculations 2 – Mark Questions

I. Write brief answer to the following questions:

Question 1.
Define a matter.
Answer:
The matter is defined as anything that has mass and occupies space. All matter is composed of atoms.

Question 2.
What is molar volume?
Answer:
Molar volume is the volume occupied by one mole of a substance in the gaseous state at STP. It is equal to 2.24 x 10-2m3 (22.4 L).

Question 3.
The approximate production of Na2CO3 per month is 424 x 106 g while that of methyl alcohol is 320 x 106 g. Which is produced more in terms of moles?
Answer:
Na2CO3 mass = 424 x 106g
Molecular mass of Na2CO3 = (23 x 2) + 12 + (16 x 3)
= 46+ 12 +48
= 106 g
No. of moles of Na2CO3 Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
= 4 x 106 moles
Methyl alcohol mass = 320 x 106 g
Molecular mass of CH3OH = 12 + (1 x 4)+ 16 = 32 g
= 12 + 4 + 16 = 32 g
No. of moles of Methyl alcohol Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
= 10 x 106 moles.
∴ Methyl alcohol is more produced in terms of moles.

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 4.
What is a mixture? Give an example.
Answer:
Mixtures consist of more than one chemical entity present without any chemical interactions. They can be further classified as homogeneous or heterogeneous mixtures based on their physical appearance.
Example: Air

Question 5.
Find the molecular mass of FeSO4.7H2O
Answer:
Sum of Atomic mass of all elements = Molecular mass
Atomic mass of Fe = 56.0
Atomic mass of S = 32.0
Atomic mass of 4[O] = 64.0
Atomic mass of 14[H] = 14.0
Atomic mass of 7[O] = 112.0 = 278.0
Molecular mass of FeSO4.7H2O = 278 g.

Question 6.
What is an element? Give examples.
Answer:
An element consists of only one type of atom. The element can exist as monoatomic or polyatomic units. The polyatomic elements are called molecules.
Example: Gold (Au), Hydrogen (H2).

Question 7.
How many moles of glucose are present in 720 g of glucose?
Answer:
Glucose = C6H4O4
Molecular mass of Glucose = (12 x 6) + (1 x 12) + (16 x 6)
= 72 + 12 + 96 = 180
\(\frac {720}{180}\) = 4 moles.

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 8.
Define atomic mass unit (AMU).
Answer:
The AMU or unified atomic mass is defined as one-twelfth of the mass of a Carbon – 12 atoms in its ground state.
i.e., 1 amu (or) 1 u ≈ 1.6605 X 10-27 kg

Question 9.
What do you understand by the terms acidity and basicity?
Answer:
Acidity: The number of hydroxyl ions present in one mole of a base is known as the acidity of the base.
Basicity: The number of replaceable hydrogen atoms present in a molecule of acid is referred to as its basicity.

Question 10.
Calculate the equivalent mass of bicarbonate ion.
Answer:
Bicarbonate ion = HCO3
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Formula mass of HCO3 = 1 + 12 + 48 = 61
Equivalent mass of HCO3 = \(\frac {61}{1}\) = 61

Question 11.
What is relative molecular mass?
Answer:
Relative molecular mass is defined as the ratio of the mass of a molecule to the unified atomic mass unit.

Question 12.
Calculate the equivalent mass of hydrated sodium carbonate.
Answer:
Hydrated sodium carbonate = Na2CO3. 10H2O
Molecular mass of Na2CO3. 10H2O = (23 x 2) + (2 x 1) + (16 x 13) + (1 x 20)
= 46 + 12 + 208 + 20 = 286
Equivalent mass of Na2CO3. 10H2O = \(\frac {Molecular mass}{Acidity}\)
= \(\frac {286}{2}\) = 143

Question 13.
What are antacids?
Answer:
Antacids are commonly used medicines for treating heartburn and acidity. Antacids contain mostly magnesium hydroxide or aluminium hydroxide that neutralizes the excess acid.

Question 14.
Boric acid, H3BO3 is a mild antiseptic and is often used as an eyewash. A sample contains 0.543 mol H3BO3. What is the mass of boric acid in the sample?
Answer:
Molecular mass of H3BO3 = (1 x 3) + (11 x 1) + (16 x 3) = 62
The boric acid sample contains 0.543 moles.
Mass of 0.543 mole of Boric acid = Molecular mass x mole
= 62 x 0.543
= 33.66 g

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 15.
A compound contains 50% of X (atomic mass 10) and 50% Y (atomic mass 20). Give its molecular formula
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
∴ The Empirical Formula is X3Y
Empirical Formula mass = 20 + 20 = 40
Molecular mass = Sum of atomic mass = 40
n = 1, Molecular formula = (Empirical Formula )n = (X2Y)1 = X2Y.

Question 16.
Define molar mass.
Answer:
Molar mass is defined as the mass of one mole of a substance. The molar mass of a compound is equal to the sum of the relative atomic masses of its constituents expressed in g mol-1.

Question 17.
Define matter. What are the types of matter?
Answer:

  • A matter is anything which has mass and occupies space.
  • Matters exist in all three states such as solid, liquid, and gas.

Question 18.
Prove that states of matter are interconvertible.
Answer:
States of matter are interconvertible by changing temperature and pressure.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 19.
What is the basicity of an acid? Give an example.
Answer:
Basicity of an acid is the number of moles of ionizable H+ ions present in 1 mole of the acid. The basicity of sulphuric acid (H2SO4) is 2.

Question 20.
Differentiate an element and an atom.
Answer:

  • An atom is the ultimate smallest electrically neutral, being made up of fundamental particles such as the proton, neutron, and electron.
  • An element consists of only one type of atom. Elements are further divided into metals, non-metals, and metalloids.

Question 21.
Distinguish between a molecule and a compound
Answer:
Molecule:

  • A molecule is the smallest particle made up of one or more than one atom in a definite ratio having stable and independent existence.
  • e.g. Na – Monoatomic molecule O2 – Diatomic molecule P4 – Poly atomic molecule

Compound:

  • A molecule which contains two or more atoms of different elements are called a compound molecule.
  • e.g. CO2 – Carbon dioxide CH4 – Methane H2O – Water

Question 22.
What is the molecular formula of a compound?
Answer:
The molecular formula of a compound is the formula written with the actual number of different atoms present in one molecule as a subscript to the atomic symbol.

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 23.
Define the molecular mass of a substance.
Answer:
Molecular mass of a substance (element or compound) represents the number of times the molecule of that substance is heavier than 1 / 12th of the mass of an atom of C-12 isotope. Molecular mass = 2 x Vapour density

Question 24.
Calculate the molecular mass of Sulphuric acid (H2SO4). Element
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 25.
What are redox reactions?
Answer:
The reaction involving loss of electron is oxidation and gain of the electron is reduced. Both these reactions take place simultaneously and are called as redox reactions.

Question 26.
Calculate the number of moles present in 60 g of ethane.
Answer:
No. of moles =Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations  = \(\frac {W}{M}\)
Molar Mass of ethane (C2H6) = 24 + 6 = 30
Number of moles in 60 g of ethane = \(\frac {60}{30}\) = 2 moles.

Question 27.
Define oxidation and reduction in terms of oxidation number.
Answer:
A reaction in which the oxidation number of the element increases is called oxidation whereas the reaction in which oxidation number decreases is called reduction.

Question 28.
Calculate the equivalent mass of (i) Sulphate ion (ii) Phosphate ion.
Answer:
(i) Sulphate ion (SO42-).
Equivalent mass of Sulphate = Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations = \(\frac {32 + 64}{2}\) = \(\frac {96}{2}\) = 48 g eq-1

(ii) Phosphate ion (P043-)
Molar mass of Phosphate ion = Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations = \(\frac {31 + 64}{3}\) = \(\frac {95}{2}\) = 31.6 = 31.6g eq-1

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 29.
Calculate the equivalent mass of sulphuric acid.
Answer:
Sulphuric acid = H2SO4
Molar mass of Sulphuric acid = 2 + 32 + 64 = 96
Basicity of Sulphuric acid = 2
Equivalent mass of acid = Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations = \(\frac {96}{2}\) = 49g eq-1

Question 30.
How many moles of hydrogen is required to produce 20 moles of ammonia?
Answer:
3H2 + N2 → 2NH3
A per stoichiometric equation,
No. of moles of hydrogen required for 2 moles of ammonia 3 moles
No. of moles of hydrogen required for 20 moles of ammonia = \(\frac {3}{2}\) x 20 = 30 moles.

Question 31.
Calculate the amount of water produced by the combustion of 32 g of methane.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
As per the stoichiometric equation,
16 g of methane produces 36 g of H2O
∴ 32 g of methane will produce = \(\frac {36}{16}\) x 32 = 72 g of water.

Question 32.
How much volume of Carbon dioxide is produced when 25 g of calcium carbonate is heated completely under standard conditions?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
100 g of CaCO3 produces 22.4 L of CO2.
∴ 25 g of CaCO3 will produce = \(\frac {22.4}{100}\) x 25 = 5.6 L of CO2.

Question 33.
How much volume of chlorine is required to prepare 89.6 L of HCl gas at STP?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
2 x 22.4 L of HCl is produced by 22.4 L of Cl2.
∴ 89.6 L of HCl will be produced by = img = 89.6 L = 44.8 L of chlorine.

Question 34.
What is meant by limiting reagent?
Answer:
A large excess of one reactant is supplied to ensure the more expensive reactant is completely converted to the desired product. The reactant used up first in a reaction is called the limiting reagent.

Question 35.
On the formation of SF6 by the direct combination of S and F2, which is the limiting reagent? Prove it.
Answer:
SF6 is formed by burning Sulphur in an atmosphere of Fluorine. Suppose 3 moles of S is allowed to react with 12 moles of Fluorine.
S(l) +3F2(g) → SF6(g)
As per the stoichiometric reaction, one mole of S reacts with 3 moles of fluorine to complete the reaction. Similarly, 3 moles of S requires only 9 moles of fluorine.
∴ It is understood that the limiting reagent is Sulphur and the excess reagent is Fluorine.

Question 36.
Mention any 4 redox reaction that takes place in our daily life.
Answer:

  1. Burning of cooking gas, wood
  2. Rusting of iron articles
  3. Electroplating
  4. Galvanic and electrolytic cells

Question 37.
Calculate the oxidation number of underlined elements in the following.

  1. KMnO4
  2. Cr2O72-

Answer:
1. KMnO4
1(+1) + x + 4 (-2) = 0
x – 7 = 0
∴ x = + 7
Oxidation state of Mn = +7.

2. Cr2O72-
2x + 7(-2) = -2
2x – 14 = – 2
2x = +l2
∴ x = + 6
Oxidation state of Cr = +6.

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 38.
If 10 volumes of H2 gas react with 5 volumes of O2 gas, how many volumes of water vapour would be produced?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Thus 2 volumes of H2 reacts with 1 volume of O2 to produce 2 volumes of H2O(g).
10 volumes of H2 would react with 5 volumes of O2 to produce 10 volumes of H2O(g)
Thus 10 volumes of H2O will be produced.

Question 39.
Which one of the following will have the largest number of atoms?

  1. 1 g of Au(s)
  2. 1 g of Na(s)
  3. 1 g of Li(s)
  4. 1 g of Cl2 (g)

Answer:
1. Molar mass of Au = 197 g mol-1.
No: of atoms in 1 g of Au = \(\frac {1}{197}\) x 6.023 x 1023

2. Molar mass of Na = 23 g mol-1.
No of atoms in 1 g of Na = \(\frac {1}{23}\) x 6.023 x 1023

3. Molar mass of Li = 7 g mol-1
No. of atoms in 1 g of Li = \(\frac {1}{7}\) x 6.023 x 1023

4.  Molar mass of Cl2 = 35.5 g mol-1
No. of atoms in 1 g of Cl2 = \(\frac {1}{35.5}\) x 6.023 x 1023

Comparing the number of atoms, the largest number of atoms will be present in 1 g of Li. Since the mass is same in each case, the element with the lowest molar mass would have the largest number of atoms.
∴ Li with lowest molar mass would have the largest number of atoms.

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 40.
Calculate the relative molecular mass of glucose. (Given: Atomic mass of C =12, H = 1.008 and O = 16).
Answer:
Relative molecular mass of glucose (C6H12O6)
= (6 × 12) + (12 × 1.008) + (6 × 16)
= 72 + 12.096 + 96
= 180.096.

Question 41.
Justify the following reaction is a redox reaction.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
In the above reaction, oxygen is removed from CuO. So CuO gets reduced. Oxygen is added to H2 to form water. So H2 gets oxidised. i.e. In CuO, oxidation number of Cu +2 is reduced to O whereas in H2, oxidation of H2 O is increased to + 1. So the above reaction is a redox reaction.

Samacheer Kalvi 11th Chemistry Basic Concepts of Chemistry and Chemical Calculations 3 – Mark Questions

Question 1.
Distinguish among the different physical states of matter.
Answer:
Differences among three physical states of matter (solid, liquid and gas) are as follows

S.No. Properties Solid Liquid Gas
1. Volume Definite Definite No definite
2. Shape Definite No definite No definite
3. Molecular arrangement Very closely packed Loosely packed Very loosely packed
4. Freedom of movement Not much freedom Move around and better than solid Move easy and fast
5. Compressibility Non-compressible Less compressible Easily compressible

Question 2.
Define equivalent mass of salt.
Answer:
Equivalent mass of a salt:
It is defined as the number of parts by mass of the salt that is produced by the neutralization of one equivalent of an acid by a base. Therefore the equivalent mass of the salt is equal to its molar mass.

Question 3.
Describe the concept of stoichiometry.
Answer:
Stoichiometry gives the numerical relationship between chemical quantities in a balanced chemical equation. By applying the concept of stoichiometry, we can calculate the number of reactants required to prepare a specific amount of a product and vice versa using a balanced chemical equation.
Consider the following chemical reaction.
C(s) + O2(g) → CO2(g)
From this equation, we learned that 1 mole of carbon reacts with 1 mole of an oxygen molecule to form 1 mole of carbon dioxide.
1 mole of C ≡ 1 mole of O2 ≡ 1 mole of CO2
The symbol ‘≡’ means ‘stoichiometrically equal to’.

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 4.
Calculate the equivalent mass of hydrated ferrous sulphate.
Answer:
Hydrated ferrous sulphate = FeSO4.7H2O
Ferrous sulphate – Reducing agent
Ferrous sulphate reacts with an oxidising agent in acid medium according to the equation.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
16 parts by mass of oxygen oxidised 304 g of FeSO4.
8 parts by mass of oxygen will oxidise \(\frac{304}{16}\) x 8 parts by mass of FeSO4.
= 152
Equivalent mass of Ferrous sulphate (Anhydrous) = 152
Equivalent mass of crystalline Ferrous sulphate FeSO4 .7H2O = 152 + 126 = 278

Question 5.
Give differences between empirical and molecular formula
Answer:
Empirical Formula:

  • The empirical formula is the simplest formula.
  • It shows the ratio of a number of atoms of different elements in one molecule of the compound.
  • It is calculated from the percentage of composition of the various elements in one molecule.
  • For example, the Empirical formula of Benzene = CH.

Molecular Formula:

  • Molecular Formula is the actual formula.
  • It shows the actual number of different types of atoms present in one molecule of the compound.
  • It is calculated from the Empirical formula Molecular Formula = (Empirical formula)n
  • Molecular formula of Benzene = C6H6.

Question 6.
A sample of hydrated copper sulphate is heated to drive off the water of crystallization, cooled and reweighed 0.869 g of CuSO4 aH2O gave a residue of 0.556 g. Find the molecular formula of hydrated copper sulphate.
Answer:
0.869 g of CuSO4.aH2O gave a residue of 0.556 g of Anhydrous CuSO4.
∴ Weight of a H2O molecule = 0.869 – 0.556 = 0.313 g
Molecular weight of H2O = (1 x 2) + 16 = 2 + 16 = 18
No. of moles of water = \(\frac{Mass}{Molecular mass}\)
CuSO4.5H2O – Molecular mass = 63.5 + 32 + 64 + 90 = 249.5 g
249.5 g of CuSO4.5H2O on heating gives 159.5 g of CuSO4.
0.869 g of CuSO4.aH2O on heating gives = \(\frac{159.5}{249.5}\) x 0.869
= 0.556 g of anhydrous CuSO4 ∴ a = 5
The molecular formula of hydrated copper sulphate = CuSO4.5H2O

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 7.
Balance by oxidation number method: Mg + HNO3 → Mg(NO3)2 + NO2 + H2O
Step – 1
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Step – 2
Mg + 2HNO3 -» Mg(NO3)2 + NO2 + H2O
Step – 3
To balance the number of oxygen atoms and hydrogen atoms 2HNO3 is multiplied by 2.
Mg + 4HNO3 → Mg(NO3)2 + 2NO2 + H2O
Step – 4.
To balance the number of hydrogen atoms, the H2O molecule is multiplied by 2.
Mg + 4HNO3 → Mg(NO3)2 + 2NO2 + 2H2O

Question 8.
Explain about the classification of matter.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 9.
Calculate the mass of the following atoms in amu,
(a) Helium (mass of He = 6.641 x 10-24 g)
(b) Silver (mass of Ag = 1.790 x 10-22 g)
1 amu = 1.66056 x 10-24
Answer:
(a) The mass of Helium atom in amu = \(\frac{6.641 \times 10^{-24}}{1.66056 \times 10^{-24}}\) = 3.9992 amu.
(b) The mass of Silver atom in amu = \(\frac{1.790 \times 10^{-22}}{1.66056 \times 10^{-24}}\) = 107.79 amu.

Question 10.
Calculate the number of atoms present in 1 Kg of gold.
Answer:
The atomic mass of Gold = 197 g mol-1.
197 g of gold contains 6.023 x 1023 atoms of gold.
∴ 1000 g of gold will contain = \(\frac{1000 \times 6.023 \times 10^{23}}{197}\)
= 3.055 x 1024 atoms of Gold.

Question 11.
Calculate the molar volume of 146 g of HCl gas and the number of molecules present in it.
Answer:
Molar mass of HCl = 36.5 g
The molar volume of 36.5 g (1 mole) of HCl = 2.24 x 102 m3.
∴ The volume of 146 g (4 moles) of HCl = \(\frac{2.24 \times 10^{-2}}{36.5}\) x 146
= 8.96 x 10 m3
No. of molecules in 146 g of HCl = 4 N
= 4 x Avogadro Number
= 4 x 6.023 x 1023
= 24.092 x 1023
= 2.4092 x 1024 molecules.

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 12.
Calculate the molar mass of 20 L of gas weighing 23.2 g at STP.
Molar mass =Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Molar volume at STP = 2.24 x 10-2 m3 = 22.4 L (or) 22400 cc.
Molar mass of the gas at STP = \(\frac{23.2 x 22.4}{20}\) = 25.984 g.

Question 13.
0.6 g of a metal gives on oxidation 1 g of its oxide. Calculate its equivalent mass.
Answer:
Mass of metal = 0.6
Mass of metal oxide = 1 g
Mass of oxygen = 1 – 0.6 = 0.4 g
0.4 g of oxygen combines with 0.6 g of metal.
∴ 8 g of oxygen will combine with = \(\frac{0.6}{0.4}\) x 8
Equivalent mass of the metal = 12 g eq-1.

Question 14.
How would you calculate the equivalent mass of anhydrous oxalic acid and hydrated oxalic acid.
Answer:
In acid medium,
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
16 g of oxygen is used for oxidation of 90 g of oxalic acid.
∴ 8 g of oxygen will oxidize = \(\frac{90}{16}\) x 8 = 45 g eq-1.
Equivalent mass of Anhydrous oxalic acid = 45 g eq-1
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
= \(\frac{126}{2}\) = 63 g eq-1.

Question 15.
A compound on decomposition in the laboratory produces 24.5 g of nitrogen and 70 g of oxygen. Calculate the empirical formula of the compound.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
the empirical formula is N2O5

Question 16.
What is the steps involve in the calculation of molecular formula from empirical formula?
Answer:
Molecular mass and empirical formula are used to deduce molecular formula of the compound.
Steps to calculate molecular formula:

  • if Empirical formula is found out from the percentage composition of elements
  • Empirical formula mass can be found from the empirical formula
  • Molecular mass is found out from the given data
  • Molecular formula = (Empirical formula)n
  • where, n = Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 17.
What is combination reaction? Give example.
When two or more substances combine to form a single substance, the reactions are called combination reactions.
A + B → C
Example:
2 Mg + O2 → 2MgO

Question 18.
What is decomposition reaction? Give two examples.
Answer:
Chemical reactions in which a compound splits up into two or more simpler substances are called decomposition reaction.
AB → A + B
Example:
2KCO3 → 2KCl + O2
PCl5 → PCl3 + Cl2

Question 19.
What is displacement reactions? Give its types. Explain with example.
Answer:
The reactions in which one ion or atom in a compound is replaced (or substituted) by an ion or atom of the other element are called displacement reactions.
AB + C → AC + B

Example:
Metal displacement
CuSO4 + Zn → ZnSO4 + Cu

Example:
Non-metal displacement
2KBr + Cl2 → 2KCl + Br2

Question 20.
What is disproportionation reactions? Give example.
Answer:
The reactions in which an element undergoes simultaneously both oxidation and reduction are called as disproportionation reactions.
Example:
P4 + 3NaOH + 3H2O → PH3 + 3NaH2PO2
2HCHO + H2O → CH3OH + HCOOH

Question 21.
What are competitive electron transfer reaction? Give example.
Answer:
These are the reactions in which redox reactions take place in different vessels and it is an indirect redox reaction. There is a competition for the release of electrons among different metals.
Example:
Zn releases electrons to Cu and Cu releases electrons to Silver and so on.
Zn(s) + Cu2+ → Zn2+(aq) + Cu(s) (Here Zn – oxidised; Cu2+ – reduced)
Cu(s) + 2Ag+ → Cu2+(aq) + 2Ag(s) (Here Cu – oxidised; Ag+ – reduced)

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 22.
Balance the following equation using oxidation number method.
Answer:
1. S + HNO3 → H2SO4 + NO2 + H2O
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
2. S + 6HNO3 → H2SO4 + NO2 + H2O
3. Balance the equation (except O and H)
S + 6HNO3 → H2SO4 + 6NO2 + H2O
4. Balance O atoms by adding 2H2O
5 + 6HNO3 → H2SO4 + 6NO2 + 2H2O

Question 23.
Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% oxygen by mass.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
the empirical formula is Fe2O3

Question 24.
In three moles of ethane (C2H6) calculate the following:

  1. Number of moles of carbon atoms.
  2. Number of moles of hydrogen atoms.
  3. Number of molecules of ethane.

Answer:

(1) 1 mole of C2H6 contains 2 moles of Carbon atoms.
∴ 3 moles of C2H6 will have 6 moles of Carbon atoms.
(2) 1 mole of C2H6 contains 6 moles of Hydrogen atoms.
∴ 3 moles of C2H6 will have 18 moles of Hydrogen atoms.
(3) 1 mole of C2H6 contains 6.023 x 1023 number of molecules.
∴ 3 moles of C2H6 will contain 3 x 6.023 x 1023molecules.

Question 25.
Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction.
4HCl(aq) + MnO2(s) → 2H2O(l) + MnCl2(aq) + Cl2(aq)
How many grams of HCl react with 5.0 g of manganese dioxide? (Atomic mass of Mn = 55 g).
Answer:
1 mole of MnO2 = 55 + 32 = 87 g.
87 g of MnO2 reacts with 4 moles of HCl. i.e. = 4 x 36.5 = 146 g of HCl.
∴ 5 g of MnO2 will react with \(\frac{146}{87}\) x 5.0 = 8.40 g.

Question 26.
The density of water at room temperature is 1.0 g/ml. How many molecules are there in a drop of water if its volume is 0.05 ml?
Answer:
Volume of drop of water = 0.05 ml
Mass of a drop of water = Volume x Density
= 0.05 ml x 1.0 g / ml.
= 0.05 g
Molar mass of water (H2O) = 18 g
18 g of water = 1 mole
0. 05 g of water = \(\frac{1}{18}\) x 0.05 = 0.0028 mol.
No. of molecules present in one mole of water = 6.023 x 1023
No. of molecules present in 0.0028 mole of water = \(\frac{6.023 \times 10^{23} \times 0.0028}{1}\) = 1.68 x 1021 water molecules

Question 27.
Balance the following equation by oxidation number method. MnO4 + Fe2+ → Mn 2+ + Fe3+ (Acidic medium)
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
MnO4 + 5e → Mn2+ ……….(1)
Fe2+ → Fe3+ + e ……….(2)
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
to balance O and H atoms H2O and H+ are added.
MnO4 + 5Fe2+ + 8H+ → 5Fe3+ + Mn2+ + 4H2O

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Samacheer Kalvi 11th Chemistry Basic Concepts of Chemistry and Chemical Calculations 5-Mark Questions

I. Answer the following questions in detail
Question 1.
Define the following (a) equivalent mass of an acid (b) equivalent mass of a base (c) equivalent mass of an oxidising agent (cl) equivalent mass of a reducing agent.
Answer:
(a) Equivalent mass of an acid:
Equivalent mass of an acid is the number of parts by mass of the acid which contains 1.008 part by mass of replaceable hydrogen atom.
Equivalent mass of an acid = Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

(b) Equivalent mass of a base:
It is defined as the number of parts by mass of the base which contains one replaceable hydroxyl ion or which completely neutralizes one gram equivalent of an acid.
Equivalent mass of a base = Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

(c) Equivalent mass of an oxidising agent:
It is defined as the number of parts by mass of an oxidising agent which can furnish 8 parts by mass of oxygen for oxidation.

(d) Equivalent mass of a reducing agent:
It is defined as the number of parts by mass of the reducing agent which is completely oxidised by 8 parts by mass of oxygen or with one equivalent of any oxidising agent.

Question 2.
Find the oxidation number of the following:
(i) S in H2SO4
⇒ 2(+1) + x + 4(-2) = 0
⇒ 2 + x – 8 = 0
⇒ x = +6
Oxidation number of S in sulphuric acid is +6.

(ii) Cr in Cr2O72-
⇒ 2x + 7(-2) =0
⇒ 2x – 14 = 0
⇒ x = + 7
Oxidation number of Cr in dicghromate ion is +7.

(iii) C in CH2F2
⇒ x + 2(+ 1) + 2(-1) = 0
⇒ x + 2 – 2 = 0
⇒ x = 0
Oxidation number of C in CH2F2 is 0.

(iv) S in SO2
⇒ x + 2(- 2) = 0
⇒ x – 4 = 0
⇒ x = + 4
Oxidation number of S in Sulphur dioxide is +4.

(v) O in OF2
⇒ x + 2(-1) = 0
⇒ x – 2 = 0
⇒ x = +2
Oxidation number of oxygen in OF2 is + 2

Question 3.
Determine the empirical formula of a compound containing K = 24.15%, Mn = 34.77% and rest is oxygen.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
empirical formula of a compound = KMnO4

Question 4.
Write the steps to be followed for writing empirical formula.
Answer:
Empirical formula shows the ratio of number of atoms of different elements in one molecule of the compound.
Steps for finding the Empirical formula:
The percentage of the elements in the compound is determined by suitable methods and from the data collected; the empirical formula is determined by the following steps.

  1. Divide the percentage of each element by its atomic mass. This will give the relative number of atoms of various elements present in the compound.
  2. Divide the atom value obtained in the above step by the smallest of them so as to get a simple ratio of atoms of various elements.
  3. Multiply the figures so obtained, by a suitable integer if necessary in order to obtain whole number ratio.
  4. Finally write down the symbols of the various elements side by side and put the above numbers as the subscripts to the lower right hand of each symbol. This will represent the empirical formula of the compound.
  5. Percentage of Oxygen = 100 – Sum of the percentage masses of all the given elements.

Question 5.
An organic compound was found to contain carbon = 40.65%, hydrogen = 8.55% and Nitrogen = 23.7%. Its vapour density was found to be 29.5. What is the molecular formula of the compound?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Empirical formula = C2H5NO
Molecular mass = 2 x vapour density = 2 x 29.5 = 59.0
Empirical formula mass of C4H5NO = 24 + 5 + 14 + 16 = 59
n = Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations  = \(\frac {59}{59}\) = 1
Empirical formula mass 59
.’. Molecular formula = (Empirical formula)n
= (C2H5NO)1
Molecular formula = C2H5NO

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 6.
Calculate the empirical and molecular formula of a compound containing 32% carbon, 4% hydrogen and rest oxygen. Its vapour density is 75.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Empirical formula = C2H3O3
Empirical formula mass = 24 + 3 + 48 = 75
Molecular mass = 2 x Vapour density = 2 x 75 = 150
n = Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations  = \(\frac {150}{75}\) = 2
Molecular formula = (Empirical formula)n
Molecular formula = C2H3O3 x 2
Molecular formula = C4H6O6

Question 7.
Explain the different types of redox reactions with example.
Answer:
Redox reactions are classified into the following types:
(1) Combination reactions:
When two or more substances combine to form a single substance, the reactions are called combination reactions.
Example:
2Mg + O2 → 2MgO

(2) Decomposition reactions:
Chemical reactions in which a compound splits up into two or more simpler substances are called decomposition reaction.
Example:
2KClO3 → 2KCl + 3O2

(3) Displacement reactions:
The reactions in which one ion or atom in a compound is replaced by an ion or atom of the other element are called displacement reactions.
Example:
CuSO4 + Zn → ZnSO4 + Cu

(4) Disproportionation reactions:
The reactions in which an element undergoes simultaneously both oxidation and reduction are called as disproportionation reactions.
Example:
2HCHO + H2O → CH3OH + HCOOH

(5) Competitive Electron transfer reactions:
These are the reactions in which redox reactions take place in different vessels and it is an indirect redox reaction. There is a competition for the release of electrons among different metals.
Example:
Zn(s) + Cu2+ → Zn2+(aq) + Cu(s)  Here Zn – oxidised; Cu2+ reduced

Question 8.
Write the steps to be followed while balancing redox equation by oxidation number method.
Answer:
Oxidation number method:
This method is based on the fact that
Number of electrons lost by atoms = Number of electrons gained by atoms

Steps to be followed while balancing Redox reactions by Oxidation Number method:

  1. Write skeleton equation representing redox reaction
  2. Write the oxidation number of atoms undergoing oxidation and reduction.
  3. Calculate the increase or decrease in oxidation numbers per atom.
  4. Make increase in oxidation number equal to decrease in oxidation number by multiplying the formula of oxidant and reductant by suitable numbers.
  5. Balance the equation atomically on both sides except O and H atoms.
  6. Balance oxygen atoms by adding required number of water molecules to the side deficient
    in oxygen atoms.
  7. Add required number of H+ ions to the side deficient in hydrogen atom if the reaction is in acidic medium.
  8. For reactions in basic medium, add H2O molecules to the side deficient in hydrogen atoms and simultaneously add equal number of OH ions on the other side of the equation.
  9. Finally balance the equation by cancelling common species present on both sides of the equation.

Question 9.
Balance the following equation by oxidation number method:
Answer:
K2Cr2O7 + KCl + H2SO4 → KHSO4 + CrO2Cl 2 + H2O
Step – 1
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Step – 2
K2Cr2O7 + KCl + H2SO4 → KHSO4 + 2CrO2C2 + H2O

Step 3.
To balance Cl atom, KC1 is multiplied by 4
K2Cr2O7 + 4KCl + H2SO4 → 2CrO2Cl2 + KHSO2 + H2O

Step 4.
To balance K atom, KHSO4 is multiplied by 6.
K2Cr2O7 + 4KCl + H2SO4 → 2CrO2Cl2 + 6KHSO4+ H2O

Step 5.
To balance O and H atoms, HSO4 is multiplied by 6, H20 is multiplied by 3.
Answer:
K2Cr2O7 + 4KCl + 6H2SO4 → 2CrO2Cl2 + 6KHSO4 + 3H2O

(2) P + HNO3 → H3PO4 + NO2 + H2O
Step – 1
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Step 2.
P + 5HNO3 → H3PO4 + 5NO2 + H2O

(3) CuO + NH3 → Cu + N2 + H2O
Step 1.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Step 2.
3CuO + 2NH3 → 3Cu + N2 + H2O

Step 3.
To balance O and N, water is multiplied by 3.
3CuO + 2NH3 → 3Cu + N2 + 3H2O

(4) Zn + HNO3 →Zn(N03)2 + NH4NO3 + H2O
Step – 1.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Step 2.
4Zn + HNO3 → 4Zn(NO3)3 + NH4NO2 + H2O

Step 3.
To balance N, HNO3 is multiplied by 10
4Zn + 10HNO3 → 4Zn(NO3)2 + NH4NO3 + H3O

Step 4.
To balance oxygen, H2O is multiplied by 3
4Zn + 10HNO3 → 4Zn(NO3)2 + NH4NO3 + 3H3O

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 10.
Balance the following equation by ion-electron method In acidic medium.
Answer:
(i) S2O32- + I2 → S2O42- + SO2 + I
Oxidation half reaction:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Reduction half reaction:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
To balance, SO2 is added on RHS of the equation.
S2O32- + I2 → S2O42- + 2I + SO2
To balance oxygen atom, S2O32- and SO2 is multiplied by 2.
2S2O32- + I2 → S2O42- + 2I – + 2SO2

(2) Sb3+ + MnO4 → Sb5+ + Mn2+
Oxidation half reaction:
Sb3+ → Sb5+ + 2e ……….(1)
Reduction half reaction:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
In equation (2), H2O is added on L.HS to balance oxygen atom.
MnO2 + 5e → Mn2+ + 4H2O ………(3)
To balance Hydrogen atoms, H’ is added on RHS.
MnO4 + 5e + 8H+ → Mn2+ + 4H2O ………(4)
Equation (4) is multiplied by 2 and equation (I) is multiplied by 5 to equalise the electrons gained and electrons lost.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

(3) MnO4 + I → MnO2 + I2
Oxidation half reaction:
2I + 2e → I ………..(1)
Reduction half reaction:
MnO4  → MnO2 + e ………..(2)                                    
Equation (1)is multiplied by 3
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Equation (2)is multiplied by 2.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
To balance oxygen and hydrogen atoms, H+ is added on RHS and H2O is added on LHS.
2MnO4 + 6I + 4H+ → 2MnO2 + 2I2 + 2H2O

In acidic medium
(4) MnO4 + Fe2+ → Mn2+ + Fe3+
Oxidation half reaction:
Fe2+ → Fe3+ + e ……..(1)
Reduction half reaction:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
To balance oxygen, H+ is added on RHS and H2O is added on LHS.
MnO4 + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O

(5) Cr(OH)4 + H2O2 → CrO4
Oxidation half reaction:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Reduction half reaction:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
To balance oxygen and hydrogen atoms, OH and H2O are added.
2Cr(OH)4 + 3H2O2 + 20H → 2CrO2 + 8H2O

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 11.
(a) Define equivalent mass of an oxidising agent.
(b) How would you calculate the equivalent mass of potassium permanganate?
(a) The equivalent mass of an oxidizing agent is the number of parts by mass which can furnish 8 parts by mass of oxygen for oxidation.
(b) Potassium permanganate is an oxidizing agent.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
80 parts by mass of oxygen are given by 316 g of KMnO4.
.’. 8 parts by mass of oxygen will be furnished by \(\frac {316}{80}\) x 8 = 31.6
Equivalent mass of KMnO4 = 31.6 g eq-1.

Question 12.
(a) Define equivalent mass of an reducing agent.
(b) How would you determine the equivalent mass of Ferrous sulphate?
Answer:
(a) The equivalent mass of a reducing agent is the number of parts by mass of the reducing agent which is completely oxidised by 8 parts by mass of oxygen or one equivalent of any oxidising agent.
(b) Ferrous sulphate is a reducing agent.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
16 parts by mass of oxygen oxidised 304 parts by mass of FeSO4.
∴ 8 parts by mass of oxygen will oxidise \(\frac {316}{80}\) x 8 parts by mass of ferrous sulphate = 152 .
The equivalent mass of ferrous sulphate (anhydrous) = 152.
The equivalent mass of crystalline ferrous sulphate is (FeSO4.7H2O) = 152 + 126 = 278
The equivalent mass of crystalline ferrous sulphate = 278.

Question 13.
A compound on analysis gave the following percentage composition: C = 24.47%, H = 4.07 %, Cl = 7 1.65%. Find out its empirical formula.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
the empirical formula is CH2Cl.

Question 14.
A laboratory analysis of an organic compound gives the following mass percentage composition: C = 60%, H = 4.48% and remaining oxygen.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
∴ the empirical formula is C9H8O4.

Question 15.
An insecticide has the following percentage composition by mass: 47.5% C, 2.54% H, and 50.0% Cl. Determine its empirical formula and molecular formulae. Molar mass of the substance is 354.5 g mol-1
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
The empirical formula is C14H9C15.
Calculation of Molecular formula:
The empirical formula mass (C14H9C15) = (14 X 12) + (9 X 1) + (5 X 35.5)
n = Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations = \(\frac {354.5}{354.5}\) = 1
Empirical formula mass 354.5
Molecular formula = (Empirical formu1a)n
= (C14H9Cl5)1
∴ Molecular formula = C14H9Cl5

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 16.
An organic fruit smelling compound on analysis has the following composition by mass: C = 54.54%, H = 9.09%, O = 36.36%. Find out the molecular formula of the compound. The vapour density of the compound was found to be 44.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
∴ The empirical formula is C2H4O
Theempirical formula mass(C2H4O) = (12 x 2)+(1 x 4)+(16 x 1) = 44
Molecular mass = 2 x Vapour density = 2 x 44 = 88
n = Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations = \(\frac {88}{44}\) = 2
Molecular formula = (Empirical formula)n
= (C2H4O)2
∴ Molecular formula = C2H4O2

Question 17.
Calculate the percentage composition of the elements present in magnesium carbonate. How many Kg of CO2 can be obtained from 100 Kg of is 90% pure magnesium carbonate.
Molar mass of MgCO3 = 84.32 g mol-1
Percentage of Mg = \(\frac {24}{84.32}\) x 100 = 28.46%
Percentage of C = \(\frac {12}{84.32}\) x 100 = 14.23%
Percentage of O3 = \(\frac {48}{84.32}\) x 100 = 57.0%
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
84.32 g of 100% pure MgCO3 gives 44g of CO2
∴ 100 x 103g of 100% pure MgCO3 gives = \(\frac {44}{84.32}\) x 100 x 103
= 52.182 x 103 g CO2
100% pure MgCO3 gives 52.182 x 103 g CO2
∴ 90% pure MgCO3 will give \(\frac{52.182 \times 10^{3}}{100}\) x 90 = 46963.8 g CO2

Question 18.
Urea is prepared by the reaction between ammonia and carbon dioxide.
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
In one process, 637.2 g of NH3 are allowed to react with 1142 g of CO2.
(a) Which of the two reactants is the limiting reagent?
(b) Calculate the mass of(NH4)2CO formed.
(c) how much of the excess reagent in grams is left at the end of the reaction?
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
No. of moles of ammonia = \(\frac{637.2}{17}\) = 37.45 mole
No. of moles of CO2 = \(\frac{1142}{44}\) = 25.95 moles
As per the balanced equation, one mole of CO2 requires 2 moles of ammonia.
∴ No. of moles of NH3 required to react with 25.95 moles of CO2 is = \(\frac {2}{1}\) x 25.95 = 51.90 moles.
∴ 37.45 moles of NH3 is not enough to completely react with CO2 (25.95 moles).
Hence, NH3 must be the limiting reagent, and CO2 is excess reagent.

(b) 2 moles of ammonia produce 1 mole of urea.
∴ Limiting reagent 37.45 moles of NH3 can produce \(\frac {1}{2}\) x 37.45 moles of urea.
= 18.725 moles of urea.
∴ The mass of 18.725 moles of urea = No. of moles x Molar mass
= 18.725 x 60
= 1123.5 g of urea.

(c) 2 moles of ammonia requires 1 mole of CO2.
∴ Limiting reagent 37.45 moles of NH3 will require x 37.45 moles of CO2.
= 18.725 moles of CO2.
∴ No. of moles of the excess reagent (CO2) left = 25.95 – 18.725 = 7.225
The mass of the excess reagent (CO2) left = 7.225 x 44 = 317.9 g CO2.

Question 19.
(a) Define oxidation number.
(b) What are the rules used to assign oxidation number?
Answer:
(a) Oxidation number refers to the number of charges an atom would have in a molecule or an ionic compound, if electrons were transferred completely.
(b) Rules to assign oxidation number:

  • Oxidation number of a substance in its elementary state is equal to zero (H2, Br2, Na)
  • Oxidation number of a mono-atomtic ion is equal to the charge on the Ton (Na4 =+ 1,
  • Oxidation number of hydrogen in a compound is +1 (except hydrides).
  • Oxidation number of hydrogen in metal hydrides is -1 (NaH, CaH2).
  • Oxidation number of oxygen in a compound is -2 (except OF2 and peroxides).
  • Oxidation number of oxygen in peroxides is -1 (He,, Na»,) = I.
  • Oxidation number of oxygen in fluorinated compounds is either +1 or +2.(OF2 = +2, O,F2 + 1).
  • Fluorine has an oxidation number -1 in all ìts compounds.
  • The sum of the oxidation number of all the atoms in neutral molecules is equal to Zero.
  • For all ions, the sum of the oxidation number of all atoms is equal to the charge of the ion.

Question 20.
Balance the following equation by oxidation number method.
C6H6 + O2 → CO2 + H2O
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
(ii) Balance the changes in O.N. by multiplying the oxidant and reductant by suitable numbers
2 C6H6 + 15 O2 → CO2 + H2O

(iii) Balance the equation atomically (except O and H).
2C6H6 + 15 O2 → 12 CO2 + H2O

(iv) Balance O atoms by adding one HzO molecule to the RHS for making the number of molecules of H2O to be 6.
2C6H6 + 15 O2 → 12 CO2 + 6H2O

Question 21.
Balance the following equation by oxidation number method.
KMnO4 + HCl → KCl + MnCl2 + H2O + Cl2
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

(ii) 2KMnO4 + 1OHCl → KCl + MnCl2 + H2O + Cl2

(iii) Balance the equation atomically (except O and H).
2KMnO4 + 1OHCl → 2KCl + 2MnCl2, + H2O + Cl2

(iv) Balance chlorine atoms by adding HC1 and multiplying Cl2 by 5.
2KMnO4 + 16HCl → 2KCl + 2MnCl2 + H2O + 5Cl2

(v) To balance O and H, H2O is multiplied by 8.
2KMnO4  + I6HCl → 2KCl + 2MnCl4 + 8H2O + 5Cl2

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 22.
Balance the following equation by oxidation number method.
KMnO4 + FeSO4 + H2S04 → K2SO4 + MnSO4 + Fe2(S04)3 + H2O
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
(ii) 2KMnO4 + 10FeSO4 + H2SO4 → K2SO4 + MnSO4 + Fe2(SO4)3+ H2O

(iii) Balance the equation atomically (except O and H) and sulphate ions.
2KMnO4 + 1OFeSO4 + 8H2SO4 → K2SO4 + 2MnSO4 + 5Fe4(SO4)3 + H2O

(iv) Balance O atoms by multiplying H2O by 8.
2KMnO4 + 1OFeSO4 + 8H2SO4 → K2SO4 + 2MnSO4 + 5Fe2(SO4)3 + 8H2O

Question 23.
Balancing of molecular equation in alkaline medium.
MnO2 + O2 + KOH → K2MnO4 + H2O
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

(ii) Balance the changes in O, N, by multiplying the oxidant and reductant by suitable numbers.
4MnO2 + 2O2+ KOH → K2MnO4 + H2O

(iii) Balance the equation atomically (except O and H).
4MnO2 + 2O2 + KOH → K2 MnO4 + H2O

(iv) Balance oxygen and hydrogen atoms by multiplying H20 by 4.
4MnO2 + 2O2 + 8KOH → 4K2MnO4 + 4H2O

Question 24.
Explain the steps involved in ion-electron method for balancing redox reaction.
Answer:
Ion electron method makes use of the Half reactions. Steps involved in this method are,

  1. Write the equation in the net ionic form without attempting to balance it.
  2. Write and locate the oxidation number of atoms undergoing oxidation and reduction from the knowledge of calculation of oxidation number.
  3. Write two half reactions showing oxidation and reduction separately.
  4. Balance oxygen atoms by adding required number of water molecules to the side deficient in oxygen atoms.
  5. Add required number of H+ ions to the side deficient in hydrogen atom if the reaction is in acidic medium
  6. Add electrons to whichever side is necessary to make up the difference in oxidation number.
  7. Add the two half reactions. The resulting equation is a net balanced equation.
  8. For reactions in basic medium, add H2O and hydrogen ion to balance H and O.
  9. Finally balance the equation by cancelling common species present on both sides of the equation.

Question 25.
Write balanced equation for the oxidation of Ferrous ions to Ferric ions by permanganate ions in acid solution. The permanganate ion forms Mn2+ ions under these conditions.
Answer:
Net ionic reaction:
MnO4 + Fe2+ + H+ → Mn2+ + Fe3+
Oxidation half reaction:
Fe2+ → Fe3+ + e ……….(1)
Reduction half reaction:
MnO4 + 5e → Mn2+ ……….(2)
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
To balance O and H, H+ and H2O are added.
MnO4 + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ +4H2O

Question 26.
A flask A contains 0.5 mole of oxygen gas. Another flask B contains 0.4 mole of ozone gas. Which of the two flasks contains greater number of oxygen atoms.
Answer:
Flask A:
1 mole of oxygen gas = 6.023 x 1023 molecules
∴ 0.5 mole of oxygen gas = 6.023 x 1023 x 0.5 molecules
The number of atoms in flask A = 6.023 x 1023 x 0.5 x 2 = 6.023 x 1023 atoms.

Flask B:
1 mole of ozone gas = 6.023 x 1023 molecules
0. 4 mole of ozone gas = 6.023 x 1023 x 0.4 molecules
The number of atoms in flask B = 6.023 x 1023 x 0.4 x 3 = 7.227 x 1023 atoms.
∴ Flask B contains a great number of oxygen atoms as compared to flask A.

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 27.
(a) Formulate possible compounds of ‘Cl’ in its oxidation state is:
0, – 1, + l, + 3,+ 5, + 7

(b) H2O2 act as an oxidising agent as well as reducing agent where as O3 act as only oxidizing agent. Prove it.
(a) (1) Cl oxidation number O in Cl2.
(2) Cl oxidation number -1 in HCl.
(3) Cl oxidation number +3 in HCl O2.
(4) Cl oxidation number +5 in KClO3.
(5) Cl oxidation number +7 in C12O7.

(b) In H2O2, oxidation number of oxygen is -1 and it can vary from O to -2 (+ 2 is possible in OF2). The oxidation number can decrease or increase, because of this H2O2 can act both oxidising and reducing agent. Ozone (O3 ) only acts as oxidising agent since it decomposes to give nascent oxygen.

Question 28.
The Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+, MnO2 and H ion. Write a balanced ionic equation for the reaction.
The skeletal equation is:
Mn3+(aq) → Mn2+(aq) + MnO2(aq) + H+(aq)
Oxidation half equation:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Balance O.N. by adding electrons,
Mn3+(aq) →  MnO2(s) + e
Balance charge by adding 4H ions,
Mn3+(aq) →  MnO2(s) + 4H+(aq) + e
Balance O atoms by adding 2H2O
Mn3+(aq) + 22H2O(l) → MnO2(s) + 4H+(aq) + e ……….(1)
Reduction half equation:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Balance ON. by adding electrons:
Mn3+(aq) + e → Mn2+(aq) …………(2)
Adding Equation (1) and (2), the balanced equation for the disproportionation reaction is
2MH3+(aq) + 2H2O(l) → MnO2(s) + Mn2+(aq) + H+(aq)

Question 29.
Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess chlorine is removed by treating with sulphur dioxide. Present a balanced equation for the reaction for this redox change taking place in water.
The skeletal equation is:
Cl2(aq) + SO2(aq) + H2O(l) → Cl(aq) + SO42-(aq)
Reduction half equation:
Cl2(aq) → Cl2-(aq)
Balance Cl atoms,
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Balance O.N. by adding electrons
Cl2(aq) + 2e → 2Cl + 2e
Oxidation half equation:
Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations
Balance O,N. by adding electrons
SO2(aq) → SO42-(aq) + 2e
Balance charge by adding 4H+ ions:
SO2(aq) → SO42-(aq) + 4H+(aq) + 2e
Balance O atoms by adding 2H2O
SO2(aq) + 2H2O(l) → SO42-(aq) + 4H+(aq) + 2e
Adding equation (1) and (2), we have,
Cl2(aq) + SO2(aq) + 2H2O(l) → Cl2(aq) + SO42-(aq) + 4H+(aq)
This represents the balanced redox reaction.

Samacheer Kalvi 11th Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Students can Download Chemistry Chapter 6 Gaseous State Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Chemistry Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Samacheer Kalvi 11th Chemistry Gaseous State Textual Evaluation Solved

I. Choose the correct answer from the following:

Question 1.
Gases deviate from ideal behavior at high pressure. Which of the following statement (s) is correct for non – ideality?
(a) at high pressure the collision between the gas molecule become enormous
(b) at high pressure the gas molecules move only in one direction
(c) at high pressure, the volume of gas become insignificant
(d) at high pressure the inter molecular interactions become significant
Answer:
(d) at high pressure the inter molecular interactions become significant

Question 2.
Rate of diffusion of a gas is …………
(a) directly proportional to its density
(b) directly proportional to its molecular weight
(c) directly proportional to its square root of its molecular weight
(d) inversely proportional to the square root of its molecular weight
Answer:
(d) inversely proportional to the square root of its molecular weight

Question 3.
Which of the following is the correct expression for the equation of state of van der Waals gas?
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 4.
When an ideal gas undergoes unrestrained expansion, no cooling occurs because the molecules ………….
(a) are above inversion temperature
(b) exert no attractive forces on each other
(c) do work equal to the loss in kinetic energy
(d) colide without loss of energy
Answer:
(b) exert no attractive forces on each other

Question 5.
Equal weights of methane and oxygen is mixed in an empty container at 298 K. The fraction of total pressure exerted by oxygen ………..
(a) 1/3
(b) 1/2
(c) 2/3
(d) 1/3 × 273 × 298
Answer:
(a) 1/3
Hint:
mass of methane = mass of oxygen = a
number of moles of methane = \(\frac {a}{16}\)
number of moles of Oxygen = \(\frac {a}{32}\)
mole fraction of Oxygen = Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
Partial pressure of oxygen = mole fraction x Total Pressure = \(\frac {1}{3}\)P

For an ideal gas, the partial pressure formula relationship is called Henry’s Law.

Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 6.
The temperatures at which real gases obey the ideal gas laws over a wide range of pressure is called …………
(a) Critical temperature
(b) Boyle temperature
(c) Inversion temperature
(d) Reduced temperature
Answer:
(b) Boyle temperature
Hint:
The temperature at which real gases obey the ideal gas laws over a wide range of pressure is called Boyle temperature

Question 7.
In a closed room of 1000 m3 a perfume bottle is opened up. The room develops a smell. This is due to which property of gases?
(a) Viscosity
(b) Density
(c) Diffusion
(d) None
Answer:
(c) Diffusion

Question 8.
A bottle of ammonia and a bottle of HCl connected through a long tube are opened simultaneously at both ends. The white ammonium chloride ring first formed will be ………….
(a) At the center of the tube
(b) Near the hydrogen chloride bottle
(c) Near the ammonia bottle
(d) Throughout the length of the tube
Answer:
(b) Near the hydrogen chloride bottle
Hint:
Rate of diffusion α 1/√m
mNH3  = 17
mHCl = 36.5
γNH3  > γHCl
Hence white fumes first formed near hydrogen chloride.

Question 9.
The value of universal gas constant depends upon ………..
(a) Temperature of the gas
(b) Volume of the gas
(c) Number of moles of the gas
(d) units of Pressure and volume
Answer:
(d) units of Pressure and volume

Question 10.
The value of the gas constant R is …………
(a) 0.082 dm3 atm.
(b) 0.987 cal mol-1 K-1
(c) 8.3 J mol-1K-1
(d) 8 erg mol-1K-1
Answer:
(c) 8.3 J mol-1K-1

Question 11.
Use of hot air balloon in sports at meteorological observation is an application of
(a) Boyle’s law
(b) Newton’s law
(c) Kelvin’s law
(d) Brown’s law
Answer:
(a) Boyle’s law

Question 12.
The table indicates the value of van der Waals constant ‘a’ in (dm3)2 atm. mol-2
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
The gas which can be most easily liquefied is ………….
(a) O2
(b) N2
(c) NH3
(d) CH4
Answer:
(c) NH3
Hint:
Higher the value of ‘a’, greater the intermolecular force of attraction, easier the liquefaction. Option (c) is correct

Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 13.
Consider the following statements.
(i) Atmospheric pressure is less at the top of a mountain than at sea level
(ii) Gases are much more compressible than solids or liquids
(iii) When the atmospheric pressure increases the height of the mercury column rises Select the correct statement.
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (i), (ii) and (iii)
Answer:
(d) (i), (ii) and (iii)

Question 14.
Compressibility factor for CO2 at 400 K and 71.0 bar is 0.8697. The molar volume of CO2 under these conditions is ………..
(a) 22.04 dm3
(b) 2.24 dm3
(c) 0.41 dm3
(d) 19.5 dm3
Answer:
(c) 0.41 dm3
Compressibility factor (z) = \(\frac {Pv}{nRT}\)
V = \(\frac {z x nRT }{p}\)
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
V = 0.41 dm3

Question 15.
If temperature and volume of an ideal gas is increased to twice its values, the initial pressure P becomes ………….
(a) 4P
(b) 2P
(c) P
(d) 3P
Answer:
(c) P
Hint:
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
P2 = P2 Option (c)

Question 16.
At identical temperature and pressure, the rate of diffusion of hydrogen gas is 3\(\sqrt{3}\) times that of a hydrocarbon having molecular formula What is the value of n?
(a) 8
(b) 4
(c) 3
(d) 1
Answer:
(b) 4.
Hint:
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
Squaring on both sides and rearranging
27 x 2 = mCnH2n-2
54 = n(12) + (2n-2)(l)
54 = 12n+2n – 2
54 = 14n – 2
n = (54 + 2)/14 = 56/14 = 4

Question 17.
Equal moles of hydrogen and oxygen gases are placed in a container, with a pin-hole through which both can escape what fraction of oxygen escapes in the time required for one-half of the hydrogen to escape. (NEET phase 1)
(a) 3/8
(b) 1/2
(c) 1/8
(d) 1/4
Answer:
(c) 1/8
Hint:
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
The fraction of oxygen that escapes in the time required for one half of the hydrogen to escape is 1/8

Question 18
The variation of volume V, with temperature T, keeping pressure constant is called the coefficient of thermal expansion ie α = \(\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_{P}\)For an ideal gas α is equal to ………..
(a) T
(b) 1/T
(c) P
(d) none of these
Answer:
(b) 1/T
Hint:
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State-

Question 19.
Four gases P, Q, R and S have almost same values of ‘b’ but their ‘a’ values (a, b are Van der Waals Constants) are in the order Q < R < S < P. At a particular temperature, among the four gases the most easily liquefiable one is ………….
(a) P
(b) Q
(c) R
(d) S
Answer:
(a) P
Hint:
Greater the ‘a’ value, casier the liquefaction

Question 20.
Maximum deviation from ideal gas is expected from (NEET)
(a) CH4(g)
(b) NH3(g)
(c) H2(g)
(d) N2(g)
Answer:
(b) NH3(g)

Question 21.
The units of Van der Waals constants ‘b’ and ‘a’ respectively
(a) mol L-1 and L atm2 mol-1
(b) mol L and L atm mol2
(c) mol-1 L and L2 atm mol-1
(d) none of these
Answer:
(c) mol-1 L and L2 atm mol-1
Hint:
an2/V2 atm
a = atm L2/mol2 = L2 mol-2 atm
nb = L
b = L /mol = L mol-1

Question 22.
Assertion : Critical temperature of CO2 is 304 K, it can be liquefied above 304 K.
Reason : For a given mass of gas, volume is to directly proportional to pressure at constant temperature.
(a) both assertion and reason arc true and reason is the correct explanation of assertion
(b) both assertion and reason are true but reason is not the correct explanation of assertion
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(d) both assertion and reason are false
Hint:
Correct Statement: Critical temperature of CO2 is 304 K. It means that CO2 cannot be liquefied above 304 K, whatever the pressure may applied. Pressure is inversely proportional to volume.

Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 23.
What is the density of N, gas at 227°C and 5.00 atm pressure? (R = 0.082 L atm K-1 mol-1)
(a) 1.40 g/L
(b) 2.81 g/L
(c) 3.41 g/L
(d) 0.29 g/L
Answer:
(c) 3.41 g/L
Hint:
Density = \(\frac {Mass}{Volume}\)
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 24.
Which of the following diagrams correctly describes the behaviour of a fixed mass of an ideal gas ? (T is measured in K)
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
For a fixed mass of an ideal gas V α T
P α 1/V
and PV = Constant

Question 25.
25 g of each of the following gases are taken at 27°C and 600 mm Hg pressure. Which of these will have the least volume?
(a) HBr
(b) HCl
(c) HF
(d) HI
Answer:
(d) HI
Hint:
At a given temperature and pressure
Volume α number of moles
Volume α Mass / Molar mass
Volume α 28 / Molar mass
i.e. if molar mass is more , volume is less. Hence Hl has the least volume.

II. Answer these questions briefly.

Question 26.
State Boyle’s law.
Answer:
At a given temperature the volume occupied by a fixed mass of a gas is inversely proportional to its pressure.
Mathematically, Boyle’s law can be written as
V ∝ \(\frac{1}{P}\)     …………..(1)
(T and n are fixed, T-temperature, n-number of moles)
V = k × \(\frac{1}{P}\) …………(2)
k – proportionality constant
PV = k (at constant temperature and mass)

Question 27.
A balloon filled with air at room temperature and cooled to a much lower temperalure can be used as a model for Charles’ law.
Answer:
Charles’ law:

  • V T at constant P and n (or) \(\frac {V}{T}\) = Constant
  • A balloon filled with air at room temperature and cooled to a much lower temperature. the size of the balloon is reduced. Because if the temperature of the gas decreases, the volume also decreases in a direct proportion.
  • When temperature is reduced, the gas molecules inside in over slower due to decreased temperature and hence the volume decreases.

Question 28.
Name two items that can serve as a model for Gay Lussac’s law and explain.
Answer:
Gay Lussac’s law:
1. P α T at constant volume (or) = \(\frac {V}{T}\)
2. Example – 1:
You fill the car type completely full of air on the hottest day of summer. The type cannot change it shape and volume. But when winter comes, the pressure inside the lyre is reduced and the shape is also reduced. This confirms that pressure and temperature are direct related to each other.
3. Example – 2:
The egg in the bottle experiment.
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
A glass bottle is taken, inside the bottle put some pieces of cotton with fire. Then place a boiled egg (shell removed) at the top of the bottle. The temperature inside the bottle increases from the fire, rising (he pressure. By scaling the bottle with egg, the fire goes on, dropping the temperature and pressure. This causes the egg to be sucked into the bottle.
P α T is proved (or) = \(\frac{P_{1}}{V_{1}}=\frac{P_{2}}{V_{2}}\)

Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 29.
Give the mathematical expression that relates gas volume and moles. Describe in words what the mathematical expression means.
Answer:
The mathematical expression that relates gas volume and moles is Avogadro’s hypothesis. It may be expressed as
V ∝ n,
\(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\) = constant
where V1 and n1 are the volume and number of moles of a gas and V2 and n2 are a different set of values of volume and number of moles of the same gas at the same temperature and pressure.

Question 30.
What are ideal gases? In what way real gases differ from ideal gases.
Answer:

  1. Ideal gases are the gases that obey gas laws or gas equation PV = nRT.
  2. Real gases do not obey gas equation. PV = nRT.
  3. The deviation of real gases from ideal behaviour is measure in terms of a ratio of PV to nRT. This is termed as compression factor (Z). Z = \(\frac {PV}{nRT}\)
  4. For ideal gases Z = 1.
  5. For real gases Z > 1 or Z < 1. For example, at high pressure real gases have Z >1 and at intermediate pressure Z < 1.
  6.  Above the Boyle point Z> 1 for real gases and below the Boyle point, the real gases first show a decrease for Z, reaches a minimum and then increases with the increase in pressure.
  7. So, it is clear that at low pressure and high temperature, the real gases behave as ideal gases.

Question 31.
Can a Van der Waals gas with a = 0 be liquefied? Explain.
Answer:
If the van der Waals constant (a) = 0 for gas, then it behaves ideally, (i.e.,) there is no intermolecular forces of attraction. So it cannot be liquefied. Moreover,
Pc = \(\frac{a}{27 b^{2}}\)
If a = 0, then Pc = 0; therefore it cannot he liquefied.

Question 32.
Suppose there ¡s a tiny sticky area on the wan of a container of gas. Molecules hitting this area stick there permanently. Is the pressure greater or less than on the ordinary area of walls?
Answer:

  • Molecules hitting the tiny sticky area on the wall of the container of gas moves faster as they get closer to adhesive surface, but this effect is not permanent.
  • The pressure on the sticky wall is greater than on the ordinary area of walls.

Question 33.
Explain the following observations?
(a) Aerated water bottles are kept under water during summer
(b) Liquid ammonia bottle is cooled before opening the seal
(c) The type of an automobile is inflated to slightly lesser pressure in summer than in winter
(d) The size of a weather balloon becomes larger and larger as it ascends up into larger altitude
Answer:
(a) Aerated water bottles contains excess dissolved oxygen and minerals which dissolved under certain pressure and if this pressure suddenly decrease due to change in atmospheric pressure, the bottle will certainly burst with decrease the amount of dissolved oxygen in water, this tends to change the aerated water into normal water.

(b) The vapour pressure of ammonia at room temperature is very high and hence the ammonia will evaporate unless the vapour pressure is decreased. On cooling the vapour pressure decreases so that the liquid remains in the same state. Hence, the bottle is cooled ’ before opening.

(c) In Summer due to hot weather conditions, the air inside the tyre expands to large volumes due to heat as compared to winter, therefore, inflated to lesser pressure in summer.

(d) As we move to higher altitude, the atmospheric pressure decreases, therefore the balloon can J easily expand to large volume.

Question 34.
Give suitable explanation for the following facts about gases.
(a) Gases don’t settle at the bottom of a container
(b) Gases diffuse through all the space available to them and
(c) Explain with an increase in temperature
Answer:
(a) Gases by definition are the least dense state of matter. They have negligible intermolecular forces of attraction. So they are all free to roam separately. So the least dense gas particles will not sink at the bottom of a container.

(b) When a sample of a gas introduced to one part of a closed container, its molecules very quickly disperse throughout the container, this process by which molecules disperse in space in response to differences in concentration is called diffusion. For e.g., you can smell perfume in a room, because it difluses into the air totally inside the room.

(c) Diffusion is faster at higher temperature because the gas molecules have greater kinetic energy. Since heat increase the motion, then diffusion happens faster.

Question 35.
Suggest why there ¡s no hydrogen (H2) in our atmosphere. Why does the moon have no atmosphere?
Answer:
Hydrogen has a tendency to combine with oxygen from water vapour. Hence, the presence of hydrogen is negligible in the atmosphere. The gravitational pull in the moon is very less and hence, there is no atmosphere in the moon.

Question 36.
Explain whether a gas approaches ideal behaviour or deviates from ideal behaviour if –
(a) it is compressed to a smaller volume at constant temperature
(b) the temperature is raised while keeping the volume constant
(c) more gas is introduced into the same volume and at the same temperature
Answer:
(a) it a gas is compressed to a smaller volume at constant temperature, pressure is increased. At high pressure with a smaller volume, the gas deviates from ideal behaviour.

(b) If a gas temperature is raised keeping the volume constant, the pressure of the gas will increase. At high pressure, the gas deviates from ideal behaviour.

(c) if more gas is introduced into the same volume and at the same temperature, the number of moles are increasing. if the volume remains same, the increased number of moles collide with each other and kinetic energy increases and pressure decreases. At increased pressure, the gas deviates from ideal behaviour.

Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 37.
Which of the following gases would you expect to deviate from ¡deal behaviour under conditions of low-temperature FCl2, or Br2? Explain.
Answer:
Bromine has a greater tendency to deviate from ideal behavior at low temperatures. The compressibility factor tends to deviate from unity for bromine.

Question 38.
Distinguish between diffusion and effusion.
Answer:
Diffusion:

  • Diffusion is the spreading of molecules of a substance throughout a space or a second substance.
  • Diffusion refers to the ability of the gases to mix with each other.
  • E.g.. Spreading of something such as brown tea liquid spreading through the water in a tea cup.

Effusion:

  • Effusion is the escape of gas molecules through a very small hole in a membrane into an evacuated area.
  • Effusion is a ability of a gas to travel through a small pin-hole.
  • E.g., pouring out something like the soap studs bubbling out from a bucket of water.

Question 39.
Aerosol cans carry clear warning of heating of the can. Why?
Answer:
Aerosol cans carry clear warning of heating of the can. As the temperature rises, pressure in the can will increase and ambient temperatures about 120°F may lead to explosions. So aerosol cans should always be stored in dry areas where they will not be exposed to excessive temperatures. You should never throw an aerosol can onto a fire or leave it in the direct sunlight. even it is empty. This is because the pressure will build up so much that the can will burst. It is due to 2 reasons.

  • The gas pressure increases.
  • More of the liquefied propellant turns into a gas.

Question 40.
When the driver of an automobile applies brake, the passengers are pushed toward the front of the car but a helium balloon is pushed toward back of the car. Upon forward acceleration the passengers are pushed toward the front of the car. Why?
Answer:
Helium floats because it is buoyant; its molecules are lighter than the nitrogen and oxygen molecules of our atmosphere and so they rise above it. In the car, it’s the air molecules that are actually getting pulled and pushed around by gravity as the result of the accelerating frame.

That means it moves in a direction opposite to the force on the surrounding air. Normally, the air is pulled downwards due to gravity, which pushes the balloon upwards. In this case, the surrounding air in the car is pulled forward by the deceleration of the car, which pushes the helium balloon backward.

Question 41.
Would it be easier to drink water with a straw on the top of Mount Everest?
Answer:
It is difficult to drink water with a straw on the top of Mount Everest. This is because the reduced atmospheric pressure is less effective in pushing water into the straw at the top of the mountain because gravity falls off gradually with height. The air pressure falls off, there isn’t enough atmospheric pressure to push the water up in the straw all the way to the mouth.

Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 42.
Write the Van der Waals equation for a real gas. Explain the correction term for pressure and volume.
Answer:
Van der Waals equation of state for real gases is
\(\left(P+\frac{a n^{2}}{V^{2}}\right)(V-n b)\) = nRT

Correction term for pressure:
\(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\) is the pressure correction. It represents the intermolecular interaction that causes the non ideal behaviour.

Correction term for Volume:
V – nb is the volume correction. it is the effective volume occupied by real gas.

Question 43.
Derive the values of critical constants from the Van der Waals constants.
Answer:
Derivation of critical constants from the Van der Waals constants:
Van der Waals equation is,
\(\left(P+\frac{a n^{2}}{V^{2}}\right)(V-n b)\) = nRT for 1 mole
From this equation, the values of critical constant PCVC and TC arc derived in terms of a and b the Van der Waals constants.
\(\left(P+\frac{a n^{2}}{V^{2}}\right)(V – b)\) = RT ………..(1)
On expanding the equestion (1)
P V + \(\frac {a}{V}\) – pb – \(\frac{\mathrm{ab}}{\mathrm{V}^{2}}\) – RT = 0 ………(2)
Multiplying eqestion (2) by \(\frac{V^{2}}{P}\),
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
equation (3) is rearranged in the powers of V
V3 – \(\left[\frac{\mathrm{RT}}{\mathrm{P}}+\mathrm{b}\right]\) V2 + \(\frac {aV}{P}\) –
= 0 ………..(4)
The above equation (4) is an cubic equation of V, which can have three roots. At the critical point. all the three values of V are equal to the critical volume VC.
i.e. V = VC.
V – VC = O ……….(5)
(V – VC)3 = O ………(6)
(V3 – 3VCV2 + 3VC3V – VC3 = 0 ………(7)
As the equation (4) is identical with equation (7), comparing the ‘V’ ternis in (4) and (7),
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
Divide equation (11) by (10)
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
When equation (12) is substituted in (10)
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
substituting the values of Vc and Pc in equation (9)
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
Critical constant a and b can be calculated using Van der Waals Constant as follows:
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 44.
Why do astronauts have to wear protective suits when they are on the surface of moon?
Answer:
Astronauts must wear spacesuits whenever they leave a spacecraft and are exposed to the environment of the moon. On the moon, there is no air to breathe and no air pressure. Moon is extremely cold and filled with dangerous radiation. Without protection, an astronaut would quickly die in space. Spacesuits are specially designed to protect astronauts from the cold, radiation, and low pressure in space. They also provide air to breathe. Wearing a spacesuit allows an astronaut to survive and work on the moon.

Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 45.
When ammonia combines with HCl, NH4 Cl is formed as white dense fumes. Why do more funies appear near HCl?
Answer:

  1. When ammonia combines with HCl, NH4 Cl is formed as white dense fumes. The reaction takes place in neutralization between a weak base and a strong acid.
  2. The property of the gas is diffusion.
  3. Diffusion of gases Ammonia and hydrogen chloride. Concentrated ammonia solution is placed on a pad in one end of a tube and concentrated HCl on the pad at the other.
  4. After about a minute, the gases diffuses far enough to meet and a ring of solid ammonium chloride is formed near the HCl end.

Question 46.
A sample of gas at 15°C at 1 atm has a volume of 2.58 dm3. Vhen the temperature is raised to 38°C at I atm does the volume of the gas increase? if so, calculate the final volume.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
V2 = 2.78 dm3 i.e. volume increased from 2.58 dm3 to 2.78 dm3.

Question 47.
A sample of gas has a volume of 8.5 dm3 at an unknown temperature. When the sample is submerged in ice water at 0°C, its volume gets reduced to 6.37 dm3. What ¡s its initial temperature?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
T1 = 364.28 K

Question 48.
Of two samples of nitrogen gas, sample A contains 1.5 moles of nitrogen in a vessel of volume of 37.6 dm3 at 298K, and the sample B is in a vessel of volume 16.5 dm3 at 298 K. Calculate the number of moles in sample B.
Answer:
nA = 1.5 mol nB = ?
VA = 37.6 dm3 VB = 16.5 dm3
(T = 298 K constant)
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 49.
Sulphur hexafluoride is a colourless, odourless gas; calculate the pressure exerted by 1.82 moles of the gas in a steel vessel of volume 5.43 dm3 at 69.5°C, assuming ¡deal gas behaviour.
Answer:
n = 1.82 mole
V = 5.43 dm3
T = 69.5 + 273 = 342.5
P = ?
PV = nRT
P = \(\frac {nRT}{V}\)
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
P = 94.25 atm.

Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 50.
Argon is an inert gas used in light bulbs to retard the vapourlzation of the tungsten filament. A certain light bulb containing argon at 1.2 atm and 18°C is heated to 85°C at constant volume. Calculate its final pressure in atm.
Answer:
P1 = 1.2 atm
T1 = 18°C + 273 = 291 K
T2 = 85°C + 273 = 358 K
P2 = ?
\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State-

Question 51.
A small bubble rises from the bottom of a lake, where the temperature and pressure are 6°C and 4 atm. to the water surface, where the temperature is 25°C and pressure is 1 atm. Calculate the final volume in (mL) of the bubble, if its initial volume is 1.5 mL.
Answer:
T1 = 6°C + 273 = 279 K
P1 = 4 atm V1 = 1.5m
T2 = 25°C + 273 = 298 K
P2 = 1 atm V 1 = ?
\(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State-
V2 = 6.41 mol.

Question 52.
Hydrochloric acid is treated with a metal to produce hydrogen gas. Suppose a student carries out this reaction and collects a volume of 154.4 x 10-3 dm3 of a gas at a pressure of 742 mm of Hg at a temperature of 298 K. What mass of hydrogen gas (in mg) did the student collect?
Answer:
V = 154.4 x 10-3dm3
P = 742 mm of Hg
T = 298 K m = ?
n = \(\frac{PV}{RT}\) = Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State-
= 0.006 mol
n = \(\frac{PV}{RT}\)
n = \(\frac{Mass}{Molar Mass}\)
Mass = n x Molar mass
= 0.006 x 2.016
= 0.0121 g = 12.1 mg.

Question 53.
It takes 192 sec for an unknown gas to diffuse through a porous wall and 84 sec for N2 gas to effuse at the same temperature and pressure. What is the molar mass of the unknown gas?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 54.
A tank contains a mixture of 52.5 g of oxygen and 65.1 g of CO2 at 300 K the total pressure in the tank ¡s 9.21 atm. Calculate the partial pressure (in atm.) of each gas in the mixture.
Answer:
mO2 = 52.5 g
PO2 = ?
mCO2 = 65.1 g
PCO2 = ?
T = 300 K P = 9.21 atm
PO2 = XO2 x total pressure
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State-
PO2 = XO2 x Total pressure
= 0.53 x 9.21 atm = 4.88 atm
PCO2 = XCO2 x Total pressure
= 0.47 x 9.21 atm = 4.33 atm

Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 55.
A combustible gas Is stored in a metal tank at a pressure of 2.98 atm at 25 °C. The tank can withstand a maximum pressure of 12 atm after which it will explode. The building in which the tank has been stored catches fire. Now predict whether the tank will blow up first or start melting? (Melting point of the metal = 1100 K).
Answer:
T1 = 298 K;
P1 = 2.98 atm;
T2 = 1100K;
P2 = ?
\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)
P2 = \(\frac{P_{1}}{T_{1}} \times T_{2}\)
= \(\frac{2.98 \mathrm{atm}}{298 \mathrm{K}} \times 1100 \mathrm{K}\) = 11 atm
At 1100 K, the pressure of the gas inside the tank will become 11 atm. Given that tank can withstand a maximum pressure of 12 atm, the tank will start melting first.

In Text Questions – Evaluate Yourself

Question 1.
Freon-I 2, the compound widely used in the refrigerator system as coolant causes depletion of ozone layer. Now It has been replaced by eco-friendly compounds. Consider 1.5 dm3 sample of gaseous freon at a pressure of 0.3 atm. If the pressure is changed to 1.2 atm. at a constant temperature, what will be the volume of the gas increased or decreased?
Answer:
Volume of freon (V1) = 1.5 dm3
Pressure (P1) = 0.3 atm
‘T’ is constant
P2 = 1.2 atm
V2 = ?
P1V1 = P2V2
V2 = \(\frac{P_{1} V_{1}}{P_{2}}\)
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State-
= 0.375 dm3
Volume decreased from 1.5 dm3 to 0.375 dm3

Question 2.
Inside a certain automobile engine, the volume of air in a cylinder is 0.375 dm3, when the pressure is 1.05 atm. When the gas is compressed to a volume of 0.125 dm3 at the same temperature, what is the pressure of the compressed air?
Answer:
V1 = 0.375dm3
V2 = 0.125 dm3
P1 = 1.05 atm
P2 = ?
T – Constant
P1V1 = P2V2
P2 = \(\frac{P_{1} V_{1}}{V_{2}}=\frac{10.5 \times 0.375}{0.125}\)
= 3.15 atm.

Question 3.
A sample of gas has a volume of 3.8 dm3 at an unknown temperature. When the sample is submerged in ice water at 0°C, its volume gets reduced to 2.27 dm3. What is its initial temperature?
Answer:
V1 = 3.8 dm3
T2 = 0°C = 273K
T1 = ? V2 = 2.27dm3
\(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\)
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State-
T1 = 457 K

Question 4.
An athlete in a kinesiology research study has his lung volume of 7.05 dm3 during a deep inhalation. At this volume the lungs contain 0.312 mole of air. During exhalation the volume of his Jung decreases to 2.35 dm3 How many moles of air does the athlete exhale during exhalation? (assume pressure and temperature remain constant)
Answer:
V1 = 7.05 dm3
V2 = 2.35 dm3
n1 = 0.312 mol
n1 =?
‘P’ and ‘T’ are constant
\(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\)
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State-
n2 = 0.104 mol
Number of moles exhaled = 0.312 – 0.104 = 0.208 moles

Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 5.
A small bubble rises from the bottom of a lake, where the temperature and pressure are 8° C and 6.4 atm. to the water surface, where the temperature ¡s 25°C and pressure is 1 atm. Calculate the final volume in (ml) of the bubble, if its initial volume is 2.1 ml.
Answer:
T1 = 8°C = 8 + 273 = 281K
P1 = 6.4atm V1 = 2.1 mol
T2 = 25°C = 25 + 273 = 298 K
P2 = 1 atm
V2 = ?
\(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State-
V2 = 14.25 ml

Question 6.
(a) A mixture of He and O2 were used ¡n the ‘air’ tanks of underwater divers for deep dives. For a particular dive 12 dm3 of O2 at 298 K, I atm. and 46 dm3 of He, at 298 K, 1 aim. were both pumped into a 5 dm3 tank. Calculate the partial pressure of each gas and the total pressure In the tank at 298 K
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
P = 1 atm
Vtotal = 5 dm3
PO2 = XO2 x Ptotal
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
PHe = 3.54 atm

Question 6.
(b) A sample of solid KClO3 (potassium chlorate) was heated in a test tube to obtain O2 according to the reaction 2KClO3 → 2KCl(s) + 3O2 The oxygen gas was collected by downward displacement of water at 295 K. The total pressure of the mixture is 772 mm of Hg. The vapour pressure of water is 26.7 mn of Hg at 300K. What is the partial pressure of the oxygen gas?
Answer:
2KCl3(s) → 2KCl(s) 3O3(g)
Ptotal = 772 mm Hg
PH2O = 26.7 mm Hg
Ptotal = PO2 + PH2O
PO2 = Ptotal – PH2O
P1 = 26.7 mm Hg
T1 = 300 K
T2 = 295 K
P2 = ?
\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
P2 = 26.26 mm Hg
∴ PO2 = 772 – 26.26
= 745.74 mm Hg

Question 7.
A flammable hydrocarbon gas of particular volume is found to diffuse through a small hole in 1.5 minutes. Under the same conditions of temperature and pressure an equal volume of bromine vapour takes 4.73 min to diffuse through the same hole. Calculate the molar mass of the unknown gas and suggest what this gas might be, (Given that molar mass of bromine = 159.8 g/ mole)
Answer:
t1 = 1.5 minutes (gas)hydrocarbon
t2 = 4.73 minutes (gas)Bromi
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
n (12) + (2n + 2) 1 = 16 (general formula for hydrocarbon CnH2n+2)
12n + 2n + 2 = 16
14n = 16 – 2
14n = 14
n = 1
The hydrocarbon is C1H2(1) + 2 = CH4

Question 8.
Critical temperature of H2O, NH3 and CO2 are 647.4, 405.5 and 304.2 K, respectively. When we start cooling from a temperature of 700 K which will liquefy first and which will liquefy finally?
Answer:
Critical temperature of a gas is defined as the temperature above which it cannot be liquefied even at high pressures.
∴ When cooling starts from 700 K, H2O vill liquefied first, then followed by ammonia and finally carbon dioxide will liquefied.

Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

In-Text Example Problems

Question 9.
In the below figure, let us find the missing parameters [volume in (b) and pressure in (c)]
P1 = 1 atm P2 = 2 atm P3 = ? atm
V1 I dm3  V2 =? dm3 V3 = 0.25 dm3
T1 = 298 K T2 = 298 K T3 = 298 K
Solution:
According to Boyle’s law, at constant temperature for a given mass of gas at constant temperature,Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
P1V1 = P2V2 = P3V3
I atm x 1 dm3 = 2 atm x V2 = P3x 0.25 dm3
∴ 2 atm x V2 = 1 atm x 1 dm3
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
V2 = 0.5 dm3
and P3 x 0.25 dm3 = 1 atm x 1 dm3
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
P3 = 4atm

Question 10.
In the below figure, let us find the missing parameters [volume in (b) and temperature in (c)]
P1 = 1 atm P2 = 1 atm P3 = 1 atm
V1 = 0.3dm3 V2 = ?dm3 V3 = 0.15dm3
T1 = 200K T2 = 300 K T3 = ? K
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
According to Charles law,
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
T3 = 100k

Question 11.
Calculate the pressure exerted by 2 moles of sulphur hexafluoride in a steel vessel of volume 6 dm3 at 70°C assuming It is an ideal gas.
Answer:
We will use the ideal gas equation for this calculation as below:
P = \(\frac {nRT}{V}\)
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
= 9.39 atm.

Question 12.
A mixture of gases contains 4.76 mole of Ne, 0.74 mole of Ar and 2.5 mole of Xe. Calculate the partial pressure of gases, if the total pressure is 2 atm. at a fixed temperature. Solve this problem using Dalton’s law.
Answer:
PNe = XNe PTotal
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
PNe = XNe PTotal = 0.595 x 2 = 1.19 atm.
PNe = XNe PTotal = 0.093 x 2 = 0. 186 atm.
PNe = XNe PTotal = 0.312 x 2 = 0.624 atm.

Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 13.
An unknown gas diffuses at a rate of 0.5 time that of nitrogen at the same temperature and pressure. Calculate the molar mass of the unknown gas.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 14.
If a scuba diver takes a breath at the surface filling his lungs with 5.82 dm3 of air what volume will the air in his lungs occupy when he drives to a depth, where the pressure ¡s 1.92 atm. (assume temperature is constant and the pressure at the surface is exactly 1 atm.)
Solution :
Temperature = Constant
Pressure at the surface = 1 atm – P1
Pressure at the depth = 1.92 atm – P2
Vdlume of air breathing at the surface of the air = 5.82 dm3 – V1
Volume of air breathing at the depth = V2 = ?
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
V1 = 3.03 dm3
The volume of air scuba diver’s lung occupy = 3.03 dm3

Question 15.
Inside a certain automobile engine, the volume of air in a cylinder is 0.475 dm3, when the pressure is 1.05 atm. When the gas is compressed, the pressure increased to 5.65 atm. at the same temperature. What is the volume of compressed air?
Solution:
Volume of air in the cylinder 0.475 dm3 – V1
Pressure of air P1 = 1.05 atm
Increased pressure P2 = 5.65 atm
Volume of air compressed V2 = ?
P1V1 = P2V2
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
V2 = 0.08827 dm3
Compressed volume of air = 0.08 827 dm3

Samacheer Kalvi 11th Chemistry Gaseous State Additional Questions Solved

I. Choose the correct answer.

Question 1.
The approximate volume percentages of nitrogen and oxygen in the atmosphere of air are _______ and ______ respectively.
(a) 21, 68
(b) 21, 78
(c) 78, 2
(d) 80, 21
Answer:
(c) 78, 2

Question 2.
The average kinetic energy of the gas molecule is …………
(a) inversely proportional to its absolute temperature
(b) directly proportional to its absolute temperature
(c) equal to the square of its absolute temperature
(d) All of the above
Answer:
(b) directly proportional to Its absolute temperature

Question 3.
The compound widely used in the refrigerator as coolant is
(a) Freon-2
(b) Freon-12
(c) Freon-13
(d) Freon-14
Answer:
(b) Freon-12

Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 4.
At constant temperature, the pressure of the gas is reduced to one-third, the volume
(a) reduce to one-third
(b) increases by three times
(c) remaining the same
(d) cannot be predicted
Answer:
(b) increases by three times

Question 5.
Hydrogen is placed in the ______ of the periodic table.
(a) Group -1
(b) group-17
(c) group-18
(d) group-2
Answer:
(a) Group -1

Question 6.
Viscosity of a liquid is a measure of ……………
(a) repulsive forces between the liquid molecules
(b) frictional resistance
(c) intermolecular forces between the molecules
(d) none of the above
Answer:
(b) frictional resistance

Question 7.
At constant temperature for a given mass, for each degree rise in temperature, all gases expand by of their volume at 0°C.
(a) 273
(b) 298
(c) \(\frac{1}{273}\)
(d) \(\frac{1}{298}\)
Answer:
(c) \(\frac{1}{273}\)

Question 8.
In Vander Waals equation of state for a non-ideal gas the net force of attraction among the molecules is given by ………..
(a) \(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\)
(b) P + \(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\)
(c) P – \(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\)
(d) – \(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\)
Answer:
(a) \(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\)

Question 9.
‘At constant volume, the pressure of a fixed mass of a gas is directly proportional to temperature”. This statement is
(a) Charle’s law
(b) Boyle’s law
(c) Gay-Lussac’s law
(d) Dalton’s law
Answer:
(c) Gay-Lussac’s law

Question 10.
Which of the following gases will have the lowest rate of diffusion?
(a) H2
(b) N2
(c) F2
(d) O2
Answer:
(c) F2

Question 11.
A mixture of gases containing 4 moles of hydrogen and 6 moles of oxygen. The partial pressure of hydrogen, if the total pressure is 5 atm, is
(a) 10 atm
(b) 0.4 atm
(c) 20 atm
(d) 2 atm
Answer:
(d) 2 atm

Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 12.
Which of the following is a diatomic gas in nature?
(a) Oxygen
(b) Ozone
(c) Helium
(d) Radon
Answer:
(a) Oxygen

Question 13.
The property of a gas which involves the movement of the gas molecules through another gas is called
(a) effusion
(b) dissolution
(c) diffusion
(d) expansion
Answer:
(c) diffusion

Question 14.
Among the following groups which contains monoatomic gases?
(a) Group 17
(b) Group 18
(c) Group 1
(d) Group 15
Answer:
(b) Group 18

Question 15.
The rate of diffusion of a gas is inversely proportional to the
(a) square of molar mass
(b) square root of density
(c) square root of molar mass
(d) square of density
Answer:
(c) square root of molar mass

Question 16.
Which of the following gas is essential for our survival?
(a) N2
(b) H2
(c) O2
(d) He
Answer:
(c) O2

Question 17.
The deviation of real gases from ideal behavior is measured in terms of
(a) expansivity factor
(b) molar mass
(c) pressure
(d) compressibility factor
Answer:
(d) compressibility factor

Question 18.
Which of the following is not chemically inert?
(a) Helium
(b) Oxygen
(c) Argon
(d) Krypton
Answer:
(b) Oxygen

Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 19.
Match the List-I and List-II using the correct code given below the list.
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 20.
Ideal gas equation for ‘n’ moles is
(a) \(\frac{P}{R}=\frac{n T}{V}\)

(b) PV = \(\frac{n R}{T}\)

(c) \(\frac{V}{T}=\frac{n P}{R}\)

(d) \(\frac{P}{T}=\frac{n R}{V}\)
Answer:
(d) \(\frac{P}{T}=\frac{n R}{V}\)

Question 21.
The SI unit of pressure is ………..
(a) Nm-2  Kg-1
(b) Pascal
(c) bar
(d) atmosphere
Answer:
(b) Pascal

Question 22.
The volume correction introduced by Van der Waals is, Videal =
(a) V – nb
(b) V + nb
(c) \(\frac{V}{n b}\)
(d) b – nV
Answer:
(b) V + nb

Question 23.
The instrument used for measuring the atmospheric pressure is …………….
(a) lactometer
(b) barometer
(c) electrometer
(d) ammeter
Answer:
(b) barometer

Question 24.
The unit of Van der Waals constant V is
(a) atm lit mol-1
(b) atm lit2 mol-2
(c) atm lit-2 mol2
(d) atm lit-1 mol2
Answer:
(b) atm lit2 mol-2

Question 25.
Mathematical expression of Boyle’s law is ………….
(a) P1V1 = P2V2
(b) \(\frac {P}{V}\) = Constant
(c) \(\frac {V}{T}\) = Constant
(J) \(\frac {P}{T}\) = Constant
Answer:
(a) P1V1 = P2V2

Question 26.
Statement-I: If the volume of a fixed mass of a gas is reduced to half at constant temperature the gas pressure doubles.
Statement-II: If the volume is halved, the density of the gas is doubled.
(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I
(b) Statement-I and II are correct but Statement-II is not the correct explanation of Statement-I
(c) Statement-I is correct but Statement-II is wrong
(d) Statement-I is wrong but Statement-II is correct
Answer:
(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I

Question 27.
The gas used in the pressure-volume isotherm study of Andrew’s experiment is
(a) N2
(b) H2S
(c) NH3
(d) CO2
Answer:
(d) CO2

Question 28.
Which one of the following is absolute zero?
(a) 293 K
(b) 273 K
(c) – 273.15°C
(d) 0°C
Answer:
(c) – 273.15°C

Question 29.
The relationship between critical volume and Vander waals constant is
(a) Vc =\(\frac{a}{R b}\)
(b) Vc = 3b
(c) Vc = \(\frac{8 a}{27 R b}\)
(d) Vc = 8b
Answer:
(b) Vc = 3b

Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 30.
The ideal gas equation is …………..
(a) PV = RT for 1 mole
(b) P1 V1 = P2V2
(c) \(\frac {P}{T}\) = R
(d) P = P2 + P2 + P2
Answer:
(a) PV = RT for 1 mole

Question 31.
In the equation PV = nRT, which one cannot be numerically equal to R?
(a) 8.31 × 107 ergs K-1 mol-1
(b) 8.31 × 107 dynes cm K-1 mol-1
(c) 8.317J K-1 mol-1
(d) 8.317L atm K-1 mol-1
Answer:
(d) 8.317L atm K-1 mol-1

Question 32.
Mathematical expression of Graham’ s law is ……………….
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 33.
The number of moles of H2 in 0.224 L of hydrogen gas at STP (273 K, 1 atm) assuming ideal gas behavior is
(a) 1
(b) 0.1
(c) 0.01
(d) 0.001
Answer:
(c) 0.01

Question 34.
The value of compression factor Z is equal to ………….
(a) \(\frac {nRT}{PV}\)
(b) \(\frac {PV}{RT}\)
(c) PV x nRT
(d) \(\frac {PV}{nRT}\)
Answer:
(d) \(\frac {PV}{nRT}\)

Question 35.
To what temperature must a neon gas sample be heated to double its pressure, if the initial volume of gas at 75°C is decreased by 15.0% by cooling the gas?
(a) 319°C
(b) 592°C
(c) 128°C
(d) 60°C
Answer:
(a) 319°C

Question 36.
The value of critical temperature of carbon dioxide is …………
(a) 273 K
(b) 303.98 K
(c) 373 K
(d) – 80°C
Answer:
(b) 303.98 K

Question 37.
Match the List-I and List-II using the correct code given below the list.
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 38.
The rate of diffusion of gases A and B of molecular weight 36 and 64 are in the ratio
(a) 9 : 16
(b) 4 : 3
(c) 3 : 4
(d) 16 : 9
Answer:
(b) 4 : 3

Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 39.
The temperature below which a gas obey Joule Thomson effect is called …………..
(a) critical temperature
(b) standard temperature
(c) inversion temperature
(d) normal temperature
Answer:
(c) Inversion temperature

Question 40.
Which of the following is true about the gaseous state?
(a) Thermal energy = molecular interaction
(b) Thermal energy >> molecular interaction
(c) Thermal energy << molecular interaction
(d) molecular forces >> those in liquids
Answer:
(b) Thermal energy >> molecular interaction

Question 41.
The temperature produced in adiabatic process of liquefaction is …………..
(a) zero kelvin
(b) -273 K
(c) 10-4 K
(d) 104 K
Answer:
(c) 10-4 K

Question 42.
Gas deviates from ideal gas nature because molecules
(a) are colourless
(b) attract each other
(c) contain covalent bond
(d) show Brownian movement
Answer:
(b) attract each other

Question 43.
The compressibility factor for an ideal gas is …………….
(a) 1.5
(b) 2
(c) 1
(d) x
Answer:
(c) 1

Question 44.
Which of the following pair will diffuse at the same rate?
(a) CO2 and N2O
(b) CO2 and NO
(c) CO2 and CO
(d) N2O and NO
Answer:
(a) CO2 and N2O

Question 45.
What are the most favourable conditions to liquefy a gas?
(a) High temperature and low pressure
(b) Low temperature and high pressure
(c) High temperature and high pressure
(d) Low temperature and low pressure
Answer:
(b) Low temperature and high pressure

Question 46.
A person living in Shimla observed that cooking food with using pressure cooker takes more time. The reason for this observation is that at high altitude …………..
(a) pressure increases
(b) temperature decreases
(c) pressure decreases
(a) temperature decreases
Answer:
(c) pressure decreases

Question 47.
Statement-I: At constant temperature PV vs V plot for real gases is not a straight line.
Statement-II : At high pressure, all gases have Z >1, but at intermediate pressure most gases have Z <1.
(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I
(b) Statement-I and II are correct but Statement-II is not the correct explanation of Statement-I
(c) Statement-I is correct but Statement-Il is wrong
(d) Statement-I is wrong but Statement-Il is correct .
Answer:
(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I

Question 48.
Statement-I: Gases do not liquefy above their critical temperature, even on applying high press ure.
Statement-II: Above critical temperature, the molecular speed is high and intermolecular
attractions cannot hold the molecules together because they escape because of high speed.
(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I
(b) Statement-I and II are correct but Statement-II is not the correct explanation of Statement-I
(c) Statement-I is correct but Statement-II is wrong
(d) Statement-I is wrong but Statement-II is correct
Answer:
(c) Statement-I and II are correct and Statement-IIs the correct explanation of Statement-I

Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 49.
NH3 gas is liquefied more easily than N2. Hence
(a) Van der Waals constant a and b of NH3 > that of N2
(b) van der Waals constants a and b of NH3 < that of N2
(c) a(NH3) > a(N2) but b(NH3) < b(N2)
(d) a(NH3) < a(N2) but b(NH3) > b(N2)
Answer:
(c) a(NH3) > a(N2) but b(NH3) < b(N2)

Question 50.
In a closed flask of 5 liters, 1.0 g of H2 is heated from 300 to 600 K, which statement is not correct’?
(a) pressure of the gas increases
(b) the rate of the collusion increase
(c) the number of moles of gas increases
(d) the energy of gaseous molecules increases
Answer:
(c) the number of moles of gas increases

Question 51.
Match the List-I and List-II using the correct code given below the list.
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements - 49
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 52.
Consider the following statements.
(i) All the gases have higher densities than liquids and solids.
(ii) All gases occupy zero volume at absolute zero.
(iii) At very low pressure all gases exhibit ideal behaviour.
Which of the above statement is/are not correct?
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (ii) and (iii) only
Answer:
(a) (i) only

Question 53.
Which of the following gas is present maximum in atmospheric air?
(a) O2
(b) N2
(c) H2
(d) radon
Answer:
(b) N2

Question 54.
Which law is used in the process of enriching the isotope of U235 from other isotopes?
(a) Boyle’s law
(b) Dalton’s law of partial pressure
(c) Graham’s law of diffusion
(d) Charles’ law
Answer:
(c) Graham’s law of diffusion

Samacheer Kalvi 11th Chemistry Gaseous State 2 – Mark Questions

Question 1.
Define Pressure? Give its unit?
Answer:
Pressure is defined as force divided by the area to which the force is applied. The SI unit of pressure is pascal which is defined as 1 Newton per square meter (Nm-2).
Pressure = \(\frac{\text { Force }\left(\mathrm{N}(\text { or }) \mathrm{Kg} \mathrm{ms}^{-2}\right)}{\text {Area }\left(\mathrm{m}^{2}\right)}\)

Question 2.
Distinguish between a gas and a vapour.
Answer:

  • Gas: A substance that is normally in a gaseous state at ordinary temperature and pressure. e.g., Hydrogen.
  • Vapour: The gaseous form of any substance that is a liquid or solid at normal temperature and pressure. e.g., At 298 K and 1 atm, water exist as water vapour.

Question 3.
State Dalton’s law of partial pressure.
Answer:
The total pressure of a mixture of non-reacting gases is the sum of partial pressures of the gases present in the mixture” where the partial pressure of a component gas is the pressure that it would exert if it were present alone in the same volume and temperature. This is known as Dalton s law of partial pressures.
For a mixture containing three gases 1, 2, and 3 with partial pressures p1, p2, and p3 in a container with volume V, the total pressure Ptotal will be given by
Ptotal = p1 + p2 + p3

Question 4.
Define atmospheric pressure. What is its value?
Answer:

  • The pressure exerted on a unit area of Earth by the column of air above it is called atmospheric pressure.
  • The standard atmospheric pressure = 1 atm.
  • 1 atm = 760 mm Hg.

Question 5.
Define the Critical temperature (Tc) of a gas?
Answer:
Critical temperature (Tc) of a gas is defined as the temperature above which it cannot be liquefied even at high pressure.

Question 6.
Most aeroplanes cabins are artificially pressurized. Why?
Answer:
The pressure decreases with the increase in altitude because there are fewer molecules per unit volume of air. Above 9200 m (30,000 ft), for example, where most commercial aeroplanes fly, the pressure is so low that one could pass out for lack of oxygen. For this reason most aeroplanes cabins arc artificially pressurized.

Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 7.
How do you understand the PV relationship?
Answer:
The PV relationship can be understood as follows. The pressure is due to the force of the gas particles on the walls of the container. If a given amount of gas is compressed to half of its volume, the density is doubled and the number of particles hitting the unit area of the container will be doubled. Hence, the pressure would increase twofold.

Question 8.
State Charles’ law.
Answer:
Charles’ law:
For a fixed mass of a gas constant pressure, the volume is directly proportional to temperature (K).
Mathematically V – T at constant P and n. (or) \(\frac {V}{T}\) = Constant (or) \(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\) = Constant

Question 9.
A gas cylinder can withstand a pressure of 15 aim. The pressure of the cylinder is measured 12 atm at 27°C. Upto which temperature limit the cylinder will not burst?
Answer:
Cylinder will burst at that temperature when it attains the pressure of 15 atm
P1 = 12 atm
T = 27° C = (27 + 273)K = 300K
P2 = 15 atm; T = ?
\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

T2 = \(\frac{15 \times 300}{12}\) = 375 K
= (375 – 273)°C = 102°C

Question 10.
State Avogadro’s hypothesis.
Answer:
Equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules. Mathematically V ∝ n
\(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\) = Constant

Question 11.
State Charles law.
Answer:
For a fixed mass of a gas at constant pressure, the volume is directly proportional to its temperature.
\(\frac{V}{T}\) = constant (at constant pressure)

Question 12.
What are the applications of Dalton’s law of partial pressure?
Answer:
1. Physicians report the pressure of the patient’s gases in blood, analyzed by hospital lab, the values are reported as partial pressures.
Gas – Normal range
PCO2 35 – 45mm of Hg
PO2  80 – 100 mm of Hg

2. When gas is collected by downward displacement of water, the pressure of dry vapour collected is computed using Dalton’s law
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements - 50

Question 13.
How can you identify a heavy smoker with the help of Dalton’s law?
Answer:
Physicians report the pressure of the patient’s gases in blood, analyzed by hospital lab. The values are reported as partial pressures.
Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements - 51
A heavy smoker may be expected to have low O2 and huge CO2 partial pressures.

Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 14.
State Gay-Lussac law.
Answer:
At constant volume, the pressure of a fixed mass of a gas is directly proportional to temperature.
P ∝ T (at constant volume)

Question 15.
Helium diffuses more than air. Give reason.
Answer:
Take two balloons, one is filled with air and another with helium. After one day, the helium balloon was shrunk, because helium being lighter diffuses out faster than the air

Question 16.
Distinguish between diffusion and effusion.
Answer:
The property of gas that involves the movement of the gas molecules through another gas is called diffusion. Effusion is another process in which a gas escapes from a container through a very small hole.

Question 17.
What is the compression factor?
Answer:
The deviation of real gases from ideal behaviour is measured in terms of a ratio of PV to nRT.
This is termed as a compression factor.
Compression factor = Z = \(\frac {PV}{nRT}\)
For ideal gases Z = 1 at all temperatures and pressures.

Question 18.

  1. Define critical temperature.
  2. What is the critical temperature of CO2 gas?

Answer:

  1. The temperature below which a gas can be liquefied by the application of pressure is known as critical temperature.
  2. The critical temperature of CO2 gas is 303.98 K.

Question 19.

  1. Define critical pressure.
  2. What is the critical pressure of CO2 gas?

Answer:

  1. Critical temperature (PC) of a gas is defined as the minimum pressure required to liquefy.
  2. The critical pressure of CO2 is 73 atm.

Question 20.
CO2 gas cannot be liquefied at room temperature. Give reason.
Answer:
Only below the critical temperature, by the application of pressure, a gas can be liquefied. CO2 has critical temperature as 303.98 K. Room temperature means (30 + 273 K) 3O3 K. At room temperature, (critical temperature) even by applying large amount of pressure CO2 cannot be liquefied. Only below the critical temperature, it can be liquefied. At room temperature, CO2 remains as gas.

Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 21.
What is meant by the Joule-Thomson effect?
Answer:
The phenomenon of lowering of temperature when a gas is made to expand adiabatically form a region of high pressure into a region of low pressure is known as Joule-Thomson effect.

Question 22.
Define inversion temperature.
Answer:
The temperature below which a gas obey Joule-Thomson effect is called inversion temperature (Ti).
Ti = \(\frac {2a}{Rb}\)

Question 23.
State and explain Boyle’s law. Represent the law graphically.
Answer:
It states that, the pressure of a fixed mass of a gas is inversely proportional to its volume if temperature is kept constant.
P – \(\frac {1}{V}\)
PV = Constant (n and T are constant)
P1v1 = P2V2
Graphical Representation:
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 24.
Give an expression for the van der Waals equation. Give the significance of the constants used in the equation. What are their units?
Answer:
\(\left(P+\frac{n^{2} a}{V^{2}}\right)\) (V- nb) = nRT
Where n is the number of moles present and ‘a’’ b’ are known as van der Waals constants.

Significance of Van der Waals constants:
Van der Waals constant ‘a’:
‘a’ is related to the magnitude of the attractive forces among the molecules of a particular gas. Greater the value of ’a’, more will be the attractive forces.
Unit of ’a’ = L2 mol-2

Van der Waals constant ‘b’:
‘b’ determines the volume occupied by the gas molecules which depends upon size of molecule.
Unit of ‘b’ = L mol-1

Question 25.
What are ideal and real gases? Out of CO2 and NH3 gases, which is expected to show more deviation from the ideal gas behaviour?
Answer:
Ideal gas:
A gas that follows Boyle’s law, Charles’ law and Avogadro law strictly is called an ideal gas. It is assumed that intermolecular forces are not present between the molecules of an ideal gas.

Real gases:
Gases which deviate from ideal gas behaviour are known as real gases. NH3 is expected to show more deviation. Since NH3 is polar in nature and it can be liquefied easily.

Samacheer Kalvi 11th Chemistry Gaseous State 3 – Mark Questions

Question 1.
Explain the graphical representation of Boyle’s law.
Answer:
Boyle’s law states that at a given temperature the volume occupied by a fixed mass of a gas is inversely proportional to its pressure. V ∝ (T and n are fixed).
If the pressure of the gas increases, the volume will decrease and if the pressure of the gas decreases, the volume will increase. So PV = Constant.
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 2.
What are the different methods of liquefaction of gases?
Answer:
There are different methods used for the liquefaction of gases:

  1. In Linde’s method, the Joule-Thomson effect is used to get liquid air or any other gas.
  2. In Claude’s process, the gas is allowed to perform mechanical work in addition to the Joule-Thomson effect so that more cooling is produced.
  3. In the Adiabatic process, cooling is produced by removing the magnetic property of magnetic material such as gadolinium sulphate. By this method, a temperature of 10-4 K i.e., as low as 0 K can be achieved.

Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 3.
Explain Charles’ law with an experimental illustration.
Answer:
Charles’ law states that for a fixed mass of a gas at constant pressure, the volume is directly proportional to temperature (K).
V ∝ T (or) \(\frac {V}{T}\) = Constant
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Volume vs Temperature:
If a balloon is moved from an ice water bath to a boiling water bath, the gas molecules inside move faster due to increased temperature and hence the volume increases.

Question 4.
Explain the graphical representation of Charles’ law.
Answer:
1. Variation of volume of the gas sample with temperature at constant pressure.
2. Each line (iso bar) represents the variation of volume with temperature at certain pressure. The pressure increases from P1 to P5.
3.  i.e. P1<P2 < P3  <P4 < P5 . When these lines are extrapolated to zero volume, they intersect at a temperature of -273.15°C.
4. All gases are becoming liquids, if they are cooled to sufficiently low temperatures.
5. In other words, all gases occupy zero volume at absolute zero. So the volume of a gas can be measured over only a limited temperature range.
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 5.
Explain graphicl representation of Gay Lussac’s law
Answer:
Gay Lussac’s law
At constant volume, the pressure of a fixed mass of a gas is directly proportional to temperature.
P – T (or) \(\frac {P}{T}\) = Constant
It can be graphically represented as shown here:
Lines in the pressure vs temperature graph are known as isochores (constant volume) of a gas.
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 6.
Explain the graphical representation of Avogadro’s hypothesis.
Answer:
Avogadro’s hypothesis states that equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules.
V ∝ n
\(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\) = Constant.
Where V1 and are the volume and number 10- of moles of a gas and V2 and n2 are the different set of values of volume and number of moles of the same gas.

Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

A better example to illustrate Avogadro’s Number of Moles (n) hypothesis is to observe the effect of pumping more gas into a balloon. When more gas molecules (particularly CO2) are passed, the volume of the balloon increases. The pressure and temperature stay constant as the balloon inflates, so the increase in volume is due to the increase in the quantity of gas inside the balloon.

Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 7.
The vapour pressure of water at 80°C is 355.5 mm of Hg. A 100 mL vessel contains water saturated with O2 at 80°C, the total pressure being 760 mm of Hg. The contents of the vessel were pumped into a 50 mL vessel at the same temperature. What is the partial pressure of O2?
Answer:
Ptotal = PH2O + PO2
PO2 = 760 – 355.5 = 404.5
When the contents were pumped into 50 mL vessel
P1 = 404.5
V1 = 100 mL
P2 =?
V2 = 50 mL
P1V1 = P2V2
P2 = \(\frac{404.5 \times 100}{50}\) = 809 mm
Thus, PO2 in 50 mL vessel = 809 mm

Question 8.
Derive the various values of R, gas constant.
Answer:
1. For standard conditions in which P is 1 atm, volume 22.4 14 dm3 for 1 mole at 273.15 K.
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
= 0.08205 7 dm3 atm mol-1 K-1

2. Where P = 105 Pascal, V = 22.71 x 10-3 m3 for I mole of a gas at 273.15 K.
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
= 8.314 pa m3 K-1 mol-1
=8.314x 10-2bar dm3 K-1mol-1
R = 8.314 J K-1 mol-1.

Question 9.
What is meant by Boyle’s temperature (or) Boyle’s point? How is it related to the compression point?
Answer:
(1) Over a range of low pressures, the real gases can behave ideally at a particular temperature called as Boyle temperature or Boyle point.
(2) The Boyle point varies with the nature of the gas.
(3) Above the Boyle point, the compression point Z > 1 for real gases i.e. real gases show positive deviation.
(4) Below the Boyle point, the real gases first show a decrease for Z, reaches a minimum and then increase with increase in pressure. So, it is clear that at low pressure and at high temperatures, the real gases behave as ideal gases.
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 10.
State and explain Dalton’s law of partial pressure.
Answer:
John Dalton stated that “the total pressure of a mixture of non-reacting gases is the sum of partial pressures of the gases present in the mixture” where the partial pressure of a component gas is the pressure that it would exert if it were present alone in the same volume and temperature. This is known as Dalton’s law of partial pressures, i.e., for a mixture containing three gases 1, 2, and 3 with partial pressures p1, p2, and p3 in a container with volume V, the total pressure Ptotal will be given by P = p1 + p2 + p3

Samacheer Kalvi 11th Chemistry Gaseous State 5 – Mark Question

Question 1.
How CH4 He and NH3 are deviating from ideal behaviour? (or) Explain how real gases deviate from ideal behaviour.
Answer:
1. The gases which obey gas equation PV = nRT are known as ideal gases. The gases which do not obey PV = nRT are known as real gases.

2. The gas laws and the kinetic theory are based on the assumption that molecules in the gas phase occupy negligible volume (assumption 1) and that they do not exert any force on one another either attractive or repulsive (assumption 2). Gases whose behaviour is consistent with these assumptions are said to exhibit ideal behaviour.

3. The following graph shows RT plotted against P for three real gases and an ideal gas at a given temperature.

4. According to ideal gas equation, PV/RT is equal to n. Plot PV/RT versus P for ‘n’ moles oigas at 0°C. For1 mole of an ideal gas PV/ RT is equal to 1 irrespective of the pressure of the gas.

5. For real gases, we observe various deviations from ideal behaviour at high pressure. At very low pressure, all gases exhibit ideal behaviour, ie. PV/RT values all converge to n as P approaches zero.
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
6. For real gases, this is true only at moderate low pressures. (≤ 5 atm) significant variation occurs as the pressure increases. Attractive forces operate among molecules at relatively short deviation.

7. At atmospheric pressure, the molecules in a gas are far apart and attractive forces arc negligible. At high pressure, the density of the gas increases and the molecules are much closer to one another. Intermolecular forces can be significant enough to affect the motion of the molecules and the gas will not behave ideally.

Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Question 2.
Derive Van der Waals equation of state.
Answer:
1. Consider the effect of intermolecular forces on the pressure exerted by a gas form the following explanation.

2. The speed of a molecule that is moving toward the wall of a container is reduced by the attractive forces exerted by its neighbours. Hence, the measured gas pressure is Q lower than the pressure the gas would exert, if it behave ideally.
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
Where ‘a’ is the proportionality constant and depends on the nature of the gas and n and V are the number of moles and volume of the container and respectively an2/ V2 is the correction term.

3. The frequency of encounters increases with the square of the number of molecules per unit volume n2/ V2. Therefore an2/ V2 represents the intermolecular interaction that causes non-ideal behaviour.

4.  Another correction is concerned with the volume ¿ccupied by the gas molecules. ‘V’ represents the volume of the container. As every individual molecule of a real gas occupies certain volume, the effective volume V- nb which is the actually available for the gas, n is the number of moles and b is a constant of gas.

5.  Hence Van der Waals equation of state for real gases are given as \(\left(P+\frac{a n^{2}}{V^{2}}\right)\) (V – nb) = nRT Where a and b are Van der Waals constants.

Question 3.
Explain about Andrew’s experimental isotherms of CO2 gas.
Answer:
1. Andrew plotted isotherms of carbon dioxide at different temperatures. it is then proved that many real gases behave in a similar manner like CO2.

2. At a temperature of 303.98 K, CO2 remains as a gas. Below this temperature, CO2, turns into liquid CO2 at 7 3atm. It is called the critical temperature of CO2.
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

3. At 303.98 K and 73 atm pressure, CO3,, becomes a liquid but remains a gas at higher temperature.

4. Below the critical temperature, the behaviour of CO2 is different. For example, consider an isotherm of CO2 at 294.5 K, it is a gas until the point B, is reached. At B, a liquid separates along the line BC, both the liquid and gas co-exist. At C, the gas is completely condensed.

5. If the pressure is higher than at C, only the liquid is compressed so, a steep rise in pressure is observed. Thus, there exist a continuity of state.

6.  A gas below the critical temperature can be liquefied by applying pressures.

Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

Activity – 1
The table below contains the values of pressure measured at different temperatures for 1 moie of an ideal gas. Plot the values in a graph and verify the Gay Lussac’s law. [Lines in the pressure vs temperature graph are known as iso chores (constant volume) of a gas]
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
Solution:
Gay Lussac’s law at constant volume = \(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State
If temperature increases, pressure also increases.
So \(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)
Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State