Class 11

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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:

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.

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:
P_N2 = 0.76atm
K_H = 7.6 x 10⁴
x = ?
P_N2 = K_H . x
0.76 = 7.6 x 10⁴x x
x = \(\frac { 0.76 }{ 7.6\times { 10 }^{ 4 } }\) = 1 x 10^-5

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:
K_H = 8 x 10⁴
(x_N2 )_{in air} = 0.5
Total pressure = 4 atm
Partial pressure of nitrogen = Mole fraction Total pressure
= O.5 x 4 = 2

Question 6.
Which one of the following is incorrect for ideal solution?
(a) ∆H_mix = 0
(b) ∆U_mix = o
(c) ∆P = P_Observed – P_{Calculated by raoults law} = 0
(d) ∆G_mix = 0
Answer:
(d) ∆G_mix = 0
Solution:
For an ideal solution, ∆S_mix \(\neq\) 0; Hence ∆G_mix \(\neq\) 0
∴ Incorrect is ∆G_mix = 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:
P_total = P₁ + P₂
= P₁ x₁ + P₂x₂
= P₁(1 – x₂) + P₂x₂
= P₁ – P₁x₂ + P₂x₂ = P₁ – x₂(P₁ – P₂)
[∵x₁ + x₂ = 1
x₁ = 1 – x₂]

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,

Question 12.
At 100°C the vapour pressure of a solution containing 6.5g a solute in 100g water is 732mm. If K_b = 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
K_b = 0.52

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)

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:
(π₁)_{non electrolute} = (π₂)_glucose
C₁RT = C₂RT

\(\frac { 6 }{ n(30) }\) = 0.025
n = \(\frac { 6 }{ 0.025 x 30 }\) = 30
∴ Molecular formula C₈H₁₆O₈

Question 16.
The K_H 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:
K_H = 4 x 10⁴ atm,
(P_O2)_air = 0.4 atm,
(x_o2)_{in solution} = ?
air – in solution
(P_O2)_air = K_H(x_o2)_{in solution}
0.4 = 4 x 10⁴(x_o2)_{in solution}
(x_o2)_{in solution} = \(\frac { 0.4 }{ 4\times { 10 }^{ 4 } }\) = 1 x 10^-5

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:
∆H_mix is negative and show negative deviation from Raoults law.

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:

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:

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

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.

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) \(\frac { n(i – 1) }{ (1 – n) }\)

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⁺, PO₄^3-)

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:
K_f = 1.86
W₂ = 5g
∆T_f = 3.64
M₂ = 142
W₁ = 45g
ΔT_f = i x K_f

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.

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

2. Normality (N):
It is defined as the number of gram equivalents of solute in I litre of the solution.

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

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,
P_solute α x_{solute in solution}
P_solute = K_H x_{solute solution}
Here, P_solute represents the partial pressure of the gas in vapour state which is commonly called vapour pressure. X_solute in solution represents the mole fraction of solute in the solution. K_H is an empirical constant with the dimensions of pressure.

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.
P_A ∝ x _A
when x_A = 1,
then k = P°_A
(P°A = vapour pressure of pure component)
P_A = P°_A . xa
P_B = 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.
P_solution ∝ x_A
P_solution = k . x_A
x_A = 1, k = P°_solvent
P°_solution = P°_solvent – P_solution
Lowering of vapour pressure = P°_solvent – P_solution
Relative lowering of vapour pressure = \(\frac { P° – P }{ P° }\) = x_B
where x_B = Mole fraction of solute.

Question 35.
What is molal depression constant? Does it depend on nature of the solute?
Answer:
K_f = molar freezing point depression constant or cryoscopic constant.
∆T_f = K_f . m,
where
∆T_f = depression in freezing point.
m = molality of the solution
K_f = cryoscopic constant
If m = I
∆T_f = K_f
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. K_f is directly proportional to the molal concentration of the solute particles.

W_B = mass of the solute
W_A = mass of solvent
M_B = 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.

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.

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:

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:
∆T_f = K_f . m i.e. ∆T_f ∝ m

∴ 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 1SO₄^2-)
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

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:
(K_H)_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 = K_H . x_solution
750 mm Hg = 4.2 x 10^-5 mm Hg . x_solution

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 = ?
K_f = 1.86K kg mol^-1
∆T_f = k_f . m

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:
P⁰_C6H6 = 640 mm Hg
W₂ = 2.2 g (non volatile solute)
W₁ = 40 g (benzene)
P_solution = 600 mm Hg
M₂ = ?

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

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
x_H2O = \(\frac { 1.67 }{ 1.67 + 0.016 }\) = \(\frac { 1.67 }{ 1.686 }\) = 0.99
x_H2O + x_glucose = 1
0.99 + x_glucose = 1
x_glucose = 1 – 0.99 = 0.01

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:

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 K_H (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

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
P_N2 = \((\frac { 80 }{ 100 })\) x Total pressure = \(\frac { 80 }{ 100 }\) x 1 atm = 0.8 atm
P_O2 = \((\frac { 20 }{ 100 })\) x 1 = 0.2 atm
According to Henry’s Law
P_solute = K_H x _{solute in solution}
P_N2 = (K_H)_Nitrogen x Mole fraction of Nitrogen in solution

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:
P⁰_{pure benzene} = 50.71 mm Hg
P⁰_nepthalene = 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 =P⁰_benzene 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

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:
P⁰_A = 10 torr
P_solution = 9 torr
W_A = 20 g
W_B = 1 g
M_A = 200 g mol^-1
M_B = ?

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
K_b = 2.42 K kg mol^-1
∆T_b = (319.692 – 319.450) K = 0.242 K

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
∆T_f = 0.2 K
k_f = 5.12 K kg mol^-1
M₂ = ?

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
(π₂) \(\frac { { W }_{ 2 } }{ 180\times 1 }\) 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. K_f for water is 1.86 K kg mol^-1.
Solution:

Given,
∆T_f = 0.680 K
m = 0.2 m,
∆T_f (observed) = 0.680K
∆T_f (Calculated) = k_f
m = 1.86 K kg mol^-1 x 0.2 mol kg^-1 = 0.372K

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:
P_solute = K_H x_{solute in solution}
At pressure 1.5 atm, p₁ = K_H x₁ ………..(1)
At pressure 6.0 atm, p₂ = K_Hx₂ …………..(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

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 P_A° is 1.013 bar?
Solution:

In a 2% solution weight of the solute is 2g and solvent is 98g
ΔP = P_A⁰ – P_solution = 1.013 – 1.004 bar = 0.009 bar

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:
∆T_b = K_b m = K_b x W₂ x 1000/M₂ x W₁
M₂ = K_b x W₂ x 1000/∆T_b 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. K_f 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

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:

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 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

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

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.

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⁺ + SO₄^2-
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

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.

Answer:
(a) 3 4 1 2

Question 2.

Answer:
(d) 3 4 2 1

Question 3.

Answer:
(c) 2 4 1 3

Question 4.

Answer:
(a) 4 3 1 2

Question 5.

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

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

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:

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:

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:

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

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:

= \(\frac { 0.5 }{ 1.5 + 0.5 }\) = \(\frac { 0.5 }{ 2.0 }\) = 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

= \(\frac { 10 }{ 50 }\) 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 =

= \(\frac { 3 }{ 60 }\) 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:

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

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:
P_A = P_A⁰ . X_A

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

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:

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

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.

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

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.

Answer:

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.

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.

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.

Question 5.
Define molarity.
Answer:
Molarity is defined as the number of moles of solute present in 1 litre of the solution.

Question 6.
Define normality.
Answer:
Normality is defined as the number of gram equivalents of solute present in 1 litre of the solution.

Question 7.
Define formality.
Answer:
Formality (F) is defined as the number of formula weight of solute present in 1 litre of the solution.

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.

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 x_A and x_B respectively. Let the number of moles of two components A and B are n_A and n_B respectively.

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.

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.

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.

Question 13.
What is meant by ppm? Where is it used?
Answer:
1. part per million =

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,
∆T_b = T – T°
Unit of ∆T_b is K Kg mole^-1.

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.

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.
P_total = P_A + P_B

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.

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 V_a is a volume of solute and V_b 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 – ∆T_b
4. Depression in freezing point – ∆T_f

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 (T_b) than the boiling point of pure solvent (T_b°). This increase in the boiling point is known as the elevation of boiling point.

Question 29.
Define ebullioscopic constant.
Answer:
Ebullioscopic constant k_b, is equal to the elevation in boiling point for 1 molal solution.

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 = \(\frac{\text { Normal (actual) molar mass }}{\text { obseved (abnormal) molar mass }}\)

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.

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

• α_dissociation = \(\frac { i – 1 }{ n – 1 }\)
• α_association = \(\frac { (1 – i)n }{ n – 1 }\)

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:

Question 2.
What are liquid solutions ? Explain with example
Answer:

Question 3.
What are solid solution? Give example.
Answer:

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.

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.

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.

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. P_solution = P⁰_toluene + x_benzene (P⁰_benzene – P⁰_toluene)

Question 12.
Derive the relationship between the relative lowering of vapour pressure and mole fraction of the solute.
Answer:
P_solution ∝ x_A by Raoult’s law. where x_A is the mole fraction of the solvent.
P_solution = k . x_A
When
x_A = 1
k = P⁰_solvent
P⁰_solvent = partial pressure of pure solvent
P_solution = P⁰_solvent . x_A

where
x_B = mole fraction of the solute
x_A + x_B = 1
x_B = 1 – x_A

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)
∆V_mixing = O

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

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., (∆H_mixing > O) and there will be a slight increase in volume (∆V_mixing > O)

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 (∆V_mixing < 0) on mixing.

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

6. Examples – Acetone + Chloroform, Chloroform + Diethyl ether

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.

Question 18.
Show that relative lowering of vapour pressure is a colligative property.
Answer:
According to Raoult’s law,
P_solution x_A, where x_A = mole fraction of solvent.
P_solution = k . x_A, where k = proportionality constant
For a pure solvent,
Vapour pressure = P°, x_A = 1
P°_solution = k x 1 = k
Substituting P°_solvent in Raoult’s law
P_solution = P°_solvent . x_A
Relative lowenng of vapour pressure

substituting P_solution as P°x_B in the above eaquation
Relative lowering of vapour pressure

x_A + x_B = I
where x_B = mole fraction of solute. It is clear that the relative lowering of vapour pressure depends only on the mole fraction ofthe solute (x_B) 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 (T_b) than the boiling point of the pure solvent (T°_b). This increase in the boiling point is known as elevation of boiling point ∆T_b = T_b – T°_b.

Question 20.
Graphically prove that Tb ¡s greater than T°_b.
Answer:

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 T_b which is greater than T°_b. The difference between these two temperatures (T_b – T°_b) gives the elevation of boiling point.
∆T_b = T_b T°_b.

Question 21.
Derive the relationship between the elevation of boiling point and molar mass of non volatile solute.
Answer:
The elevation of boiling point ∆T_b = T_b T°_b.
∆T_b is directly proportional to the concentration of the solute particles.
∆T_b ∝ m, (m = molaLiiy)
∆T_b = k_b. m, where k_b = ebullioscopic constant

Question 22.
Define

1. freezing point
2. Depression in freezing point.

Explain with graph.
Answer:

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. ∆T_f = T⁰_f – T_f

Question 23.
Define

1. cryoscopic constant
2. ebullioscopic constant

Answer:

1. ∆T_f = k_f. m, where k_f = cryoscopic constant. If m = 1. then AT_f = k_f
k_f is defined as depression in freezing point for 1 molal solution.

2. ∆T_f = k_f. m where k_f ebullioscopic constant. If m = 1 then AT_f = k_f
k_b 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.

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.

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

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.

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.

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,

• ∆H_mixing = 0
• ∆V_mixing = 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.

• ∆H_mixing \(\quad \neq\) 0
• ∆V_mixing \(\quad \neq\) 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.

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:

• ∆H_mix = 0
• ∆V_mix = 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.
∆H_mixing\(\quad \neq\) 0 and ∆V_mixing\(\quad \neq\) 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.

Question 35.
Given below is the sketch of a plant for carrying out a process.

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

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

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.

M_B = 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.

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 (K_H). At the same temperature, hydrogen is more soluble in water than helium. Which of theni will have a higher value of K_H 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 = K_H x
where P is the pressure of the gas, x is the mole fraction of the gas in the solution and K_H is the Henry’s law constant. KH is a function of the nature oIgas.

Higher the value of K_H 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 K_H 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 ∆H_mix 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.

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.

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

W_A = weight of solvent
W_B = weight of solute
M_A = Molar mass of solvent
M_B = molar mass of solute

where n_A = number of moles of solvent
n_B = number of moles of solute.
For dilute solution, n_A > > n_B, n_A + n_B = n_A
Then,
Number of moles of solvent and solute arc


From the above equation, molar mass of the solute M_B can be calculated using the known values of W_A, W_B, M_A and the measured relative lowering of vapour pressure.

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 W_B of the non-volatile solute in W_A g of solvent, then the number of moles of ‘n’ is

where
M_B = molar mass of the solute
Substituting n value, we get

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.

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 P_total 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.

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,
P_A < P_A⁰ x_A
P_B < P_B⁰ x_B
The total vapour pressure is less than P_A + P_B
P < P_A + P_B
P < P°_A x x_A + P°_B x_B
Hence
P₁ = P_A
P₂ = P_B
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.

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.

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

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

Volume of solution = 4.3 L

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

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

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
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
P_H2S = K_H x x_H2S

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
P°₁ = 23.8 mm
W₂ = 50g
M₂ (urea) = 60g mol^-1
W₁ = 850g
M₁(Water) = 18g mol^-1
Here we have to calculate P_s
Applying Raoult’s law,

Thus, relative lowering of vapour pressure = 0.017
Substituting P° = 23.8 mm Hg

We get,
23.8 – P_s = 0.017 P_s
P_s = 23.4 mm Hg
Thus, vapour pressure of water in the solution = 23.4 mm Hg

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, ∆T_b = 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,

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

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%

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

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 P_s = 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

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°
∆T_f = K_f x m
K_f = \(\frac { 2.15 }{ 0.146 }\)
Molality of glucose solution = \(\frac { 5 }{ 180 }\) x \(\frac { 1000 }{ 100 }\) = 0.278 m
∆T_f (Glucose) = \(\frac { 2.15 }{ 0.146 }\) x 0.278 = 4.09°K
Freezing point of glucose solution = 273.15 – 4.09 = 269.06 K

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 K_b 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
∆T_b = 36.86 – 35.60 = 1.26⁰C
K_b = 2.02
Molecular mass of the solute

= 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? (K_f for water = 1.86 k kg/mol^-1) (K_b for water = 0.512 K kg/mol^-1)
Solution:

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? K_f for water = 1.86 K kg mol^-1.
Solution:
W_(solute) = 1 5.0 g
W_(solvent) = 450 g
∆T_f = T°_f – T_f = 0 – ( – 0.34) = 0.34 °C
∆T_f = k_f 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

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. 
5. For e.g.. solid in liquid means solid is the solute and liquid is the solvent.
6. 
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
    

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:
CH_4(g)  + 2O_2(g)  CO_2(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

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

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

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)

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

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 SO₄^2-, SO₃^2-, S₂O₄^2-,S₂O₆^2- is ………..
(a) SO₃^2- < SO₃^2- < S₂O₄^2- < S₂O₆^2-
(b) SO₄^2- < S₂O₄^2- < S₂O₆^2-<SO₃^2-
(c) S₂O₄^2- < SO₃^2- < S₂O₆^2- < SO₄^2-
(d) S₂O₆^2- < S₂O₄^2- < SO₄^2- < SO₃^2-
Answer:
(c) S₂O₄^2- < SO₃^2- < S₂O₆^2- < SO₄^2-

Question 17.
The equivalent mass of ferrous oxalate is ……….

Answer:

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

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) =
= \(\frac {24}{28}\) × 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.

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.

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

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.

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

= 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:

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:

∴ 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.

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

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:

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:

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.

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

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

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

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

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

1. KMnO₄ + SnCl₂  + HCl → MnCl₂  + SnCl₄  + H₂O + KCl
2. C₂O₄^2- + Cr₂ O_7 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)

Reduction half reaction: (gain of electrons)

Add H₂O to balance oxygen atoms.

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. C₂O₄^2- + Cr₂ O₇^2- → Cr ³⁺ + CO₂  (in acid medium)
Oxidation half reaction:

Reduction half reaction:

To balance oxygen atoms, H₂O is added on RHS of equation (2)
Cr₂O₇^2- + 6e^– → 2Cr³⁺ + 7 H₂O ……….(3)
To balance Hydrogen atoms, H⁺ is added on LHS of equation (1)
C₂O₄^2- + 14H⁺ → 2CO₂ + 2e^– ……..(4)
To equalize the number of electrons gained and lost, multiply the equation (4) x 3.

3. Na₂S₂O₃ + I₂ → Na₂S₄O₆ + NaI (in acid medium)
Oxidation half reaction: (Loss of electron)
Na₂S₂O₃ → Na₂S₄O₆ + 2e^2- ……..(1)
Reduction half reaction: (Gain of electron)
I₂ + 2e^2-→ 2NaI …………(2)
Adding (1) and (2)

To balance oxygen,
2Na₂S₂O₃ + I₂ → Na₂S₂O₂ + 2NaI

In acidic medium
4. Zn + NO₃^– → Zn²⁺ + NO
Half reactions are –

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

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 –
Cr₂O₇^2- + 14H⁺ +6e^– → 2 Cr³⁺ + 7H₂O
Answer:
Cr₂O₇^2- + 14H⁺ +6e^– → 2 Cr³⁺ + 7H₂O

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:

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
∴  Atomic mass
Atomic mass x = \(\frac {32}{2}\) = 16
Atomic mass y = \(\frac {24}{1}\) = 24
Atomic mass z = \(\frac {44}{0.5}\) = 88

2. Empirical formula of the compound x₄ y₂ Z₁
Molecular mass of the compound = 400
= \(\frac {400}{200}\) = 2

3. Molecular formula of the compound = x₈ y₄ z₂

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:

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,

= 15 moles of hydrogen are required.

Question 2.
Calculate the amount of water produced by the combustion of 32 g of methane.
Answer:
CH_4(g) + 2O_2(g) → CO_2(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.

Combustion of 32 g CH₄ produces

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,

As per the stoichiometric equation,
1 mole (100g) CaCO₃ on heating produces 1 mole CO₂

At STP, 1 mole of CO₂ occupies a volume of 22.7 litres
At STP, 50 g of CaCO₃ on heating produces,

= 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,

= 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

Molar mass of MgCO₃ is 84 g mol^-1.
84 g MgCO₃ contain 24 g of Magnesium.
∴ 100 MgCO₃ contain

= 28.57 g Mg i.e. percentage of magnesium = 28.57 %.
84 g MgCO₃ contain 12 g of carbon


= 14.29 g of carbon
Percentage of carbon = 14.29 %.
84 g MgCO₃ contain 48 g of oxygen
∴ 100 g MgCO₃ contains

= 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

Question 6.

(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

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 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.

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 = 
= 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.

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:

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:

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

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, MnO₂^2-, Mn²⁺
(b) MnO₂, Mn²⁺, MnO₄^2-
(c) Mn²⁺, MnO₂, MnO₄^2-
(d) Mn²⁺, MnO₄^2-, MnO₂
Answer:
(c) Mn²⁺, MnO₂, MnO₄^2-
Hint:
MnO₄^– + 8H⁺ + 5e^– → Mn²⁺ + 4H₂O (acidic medium)
MnO₄ + 4H⁺ + 3e^– → MnO₂ + 2H₂O (concentrated basic medium)
MnO₄ + e^– → MnO₄^2- (dilute basic medium)

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:

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:

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

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

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.

Answer:

Question 45.
Match the List-I and List-Il using the correct code given below the list.


Answer:

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.

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:

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

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

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

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 C_nH_2n 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

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.

Answer:

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:

∴ 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

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 ……….

Answer:

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

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

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
= 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
= 10 x 10⁶ moles.
∴ Methyl alcohol is more produced in terms of moles.

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.

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

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

Question 15.
A compound contains 50% of X (atomic mass 10) and 50% Y (atomic mass 20). Give its molecular formula
Answer:

∴ 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.

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.

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 / 12^th 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:

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 = = \(\frac {W}{M}\)
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 (SO₄^2-).
Equivalent mass of Sulphate =  = \(\frac {32 + 64}{2}\) = \(\frac {96}{2}\) = 48 g eq^-1

(ii) Phosphate ion (P0₄^3-)
Molar mass of Phosphate ion =  = \(\frac {31 + 64}{3}\) = \(\frac {95}{2}\) = 31.6 = 31.6g eq^-1

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 =  = \(\frac {96}{2}\) = 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:

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:

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:

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) +3F_2(g) → SF_6(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. Cr₂O₇^2-

Answer:
1. KMnO₄
1(+1) + x + 4 (-2) = 0
x – 7 = 0
∴ x = + 7
Oxidation state of Mn = +7.

2. Cr₂O₇^2-
2x + 7(-2) = -2
2x – 14 = – 2
2x = +l2
∴ x = + 6
Oxidation state of Cr = +6.

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:

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.

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:

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) + O_2(g) → CO_2(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’.

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.

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

Question 7.
Balance by oxidation number method: Mg + HNO₃ → Mg(NO₃)₂ + NO₂ + H₂O
Step – 1

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:

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.

Question 12.
Calculate the molar mass of 20 L of gas weighing 23.2 g at STP.
Molar mass =

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,

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

= \(\frac{126}{2}\) = 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:

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 =

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)

Question 22.
Balance the following equation using oxidation number method.
Answer:
1. S + HNO₃ → H₂SO₄ + NO₂ + H₂O

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:

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) + MnO_2(s) → 2H₂O_(l) + MnCl_2(aq) + Cl_2(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:

MnO₄^– + 5e^– → Mn²⁺ ……….(1)
Fe²⁺ → Fe³⁺ + e^– ……….(2)

to balance O and H atoms H₂O and H⁺ are added.
MnO₄^– + 5Fe²⁺ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

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 =

(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 =

(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 Cr₂O₇^2-
⇒ 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:

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:

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 =  = \(\frac {59}{59}\) = 1
Empirical formula mass 59
.’. Molecular formula = (Empirical formula)_n
= (C₂H₅NO)₁
Molecular formula = C₂H₅NO

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:

Empirical formula = C₂H₃O₃
Empirical formula mass = 24 + 3 + 48 = 75
Molecular mass = 2 x Vapour density = 2 x 75 = 150
n =  = \(\frac {150}{75}\) = 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

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

Step 2.
P + 5HNO₃ → H₃PO₄ + 5NO₂ + H₂O

(3) CuO + NH₃ → Cu + N₂ + H₂O
Step 1.

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.

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

Question 10.
Balance the following equation by ion-electron method In acidic medium.
Answer:
(i) S₂O₃^2- + I₂ → S₂O₄^2- + SO₂ + I^–
Oxidation half reaction:

Reduction half reaction:

To balance, SO₂ is added on RHS of the equation.
S₂O₃^2- + I₂ → S₂O₄^2- + 2I^– + SO₂
To balance oxygen atom, S₂O₃^2- and SO₂ is multiplied by 2.
2S₂O₃^2- + I₂ → S₂O₄^2- + 2I – + 2SO₂

(2) Sb³⁺ + MnO₄ → Sb⁵⁺ + Mn²⁺
Oxidation half reaction:
Sb³⁺ → Sb⁵⁺ + 2e ^– ……….(1)
Reduction half reaction:

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.

(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

Equation (2)is multiplied by 2.

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:

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:

Reduction half reaction:

To balance oxygen and hydrogen atoms, OH and H₂O are added.
2Cr(OH)₄ + 3H₂O₂ + 20H^– → 2CrO₂^– + 8H₂O

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.

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.

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:

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:

∴ 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:

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 =  = \(\frac {354.5}{354.5}\) = 1
Empirical formula mass 354.5
Molecular formula = (Empirical formu1a)_n
= (C₁₄H₉Cl₅)₁
∴ Molecular formula = C₁₄H₉Cl₅

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:

∴ 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 =  = \(\frac {88}{44}\) = 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%

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.

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?

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

(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

(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

Question 22.
Balance the following equation by oxidation number method.
KMnO₄ + FeSO₄ + H2S0₄ → K2SO₄ + MnSO₄ + Fe2(S0₄)₃ + H₂O
Answer:

(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

(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)

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.

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) + MnO_2(aq) + H⁺_(aq)
Oxidation half equation:

Balance O.N. by adding electrons,
Mn³⁺_(aq) →  MnO_2(s) + e^–
Balance charge by adding 4H ions,
Mn³⁺_(aq) →  MnO_2(s) + 4H⁺_(aq) + e^–
Balance O atoms by adding 2H₂O
Mn³⁺_(aq) + 22H₂O_(l) → MnO_2(s) + 4H⁺_(aq) + e^– ……….(1)
Reduction half equation:

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) → MnO_2(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:
Cl_2(aq) + SO_2(aq) + H₂O_(l) → Cl^–_(aq) + SO₄^2-_(aq)
Reduction half equation:
Cl_2(aq) → Cl^2-_(aq)
Balance Cl atoms,

Balance O.N. by adding electrons
Cl_2(aq) + 2e^– → 2Cl + 2e^–
Oxidation half equation:

Balance O,N. by adding electrons
SO_2(aq) → SO₄^2-_(aq) + 2e^–
Balance charge by adding 4H⁺ ions:
SO_2(aq) → SO₄^2-_(aq) + 4H⁺_(aq) + 2e^–
Balance O atoms by adding 2H₂O
SO_2(aq) + 2H₂O_(l) → SO₄^2-_(aq) + 4H⁺_(aq) + 2e^–
Adding equation (1) and (2), we have,
Cl_2(aq) + SO_2(aq) + 2H₂O_(l) → Cl^–_2(aq) + SO₄^2-_(aq) + 4H⁺_(aq)
This represents the balanced redox reaction.