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Samacheer Kalvi 10th Science Solutions Chapter 12 Plant Anatomy and Plant Physiology

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Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 12 Plant Anatomy and Plant Physiology

Samacheer Kalvi 10th Science Plant Anatomy and Plant Physiology Textual Evaluation Solved

I. Choose the correct answer.

Question 1.
Casparian strips are present in the ______ of the root.
(a) cortex
(b) pith
(c) pericycle
(d) endodermis.
Answer:
(d) endodermis

Question 2.
The endarch condition is the characteristic feature of:
(a) root
(b) stem
(c) leaves
(d) flower
Answer:
(b) stem

You can Download Samacheer Kalvi 10th Science Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Question 3.
The xylem and phloem arranged side by side on same radius is called ______.
(a) radial
(b) amphivasal
(c) conjoint
(d) none of these.
Answer:
(c) conjoint

Question 4.
Which is formed during anaerobic respiration?
(a) Carbohydrate
(b) Ethyl alcohol
(c) Acetyl CoA
(d) Pyruvate.
Answer:
(b) Ethyl alcohol

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Question 5.
Kreb’s cycle takes place in:
(a) chloroplast
(b) mitochondrial matrix
(c) stomata
(d) inner mitochondrial membrane
Answer:
(b) mitochondrial matrix

Question 6.
Oxygen is produced at what point during photosynthesis?
(a) when ATP is converted to ADP
(b) when CO2 is fixed
(c) when H2O is splitted
(d) All of these.
Answer:
(b) when CO2 is fixed

II. Fill in the blanks

Question 1.
Cortex lies between ______.
Answer:
Epidermis and endodermis.

Question 2.
Xylem and phloem occur on the same radius constitute a vascular bundle called ______.
Answer:
Conjoint.

Question 3.
Glycolysis takes place in ______.
Answer:
The cytoplasm of the cell.

Question 4.
The source of O2 liberated in photosynthesis is ______.
Answer:
Byproduct.

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Question 5.
________ is ATP factory of the cells.
Answer:
Mitochondria.

III. State whether the statements are true or false. Correct the false statement

Question 1.
Phloem tissue is involved in the transport of water in a plant.
Answer:
False.
Correct Statement: Phloem tissue is involved in the transport of food in plants.

Question 2.
The waxy protective covering of a plant is called cuticle.
Answer:
True.

Question 3.
In monocot, stem cambium is present in between xylem and phloem.
Answer:
False.
Correct Statement: In monocot stem, the cambium is absent in between xylem and phloem.

Question 4.
Palisade parenchyma cells occur below the upper epidermis in dicot root.
Answer:
False.
Correct Statement: Palisade parenchyma cells occur below the upper epidermis in dicot leaf.

Question 5.
Mesophyll contains chlorophyll.
Answer:
True.

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Question 6.
Anaerobic respiration produces more ATP than aerobic respiration.
Answer:
True.

IV. Match the following

Question 1.

1. Amphicribal (a) Dracaena
2. Cambium (b) Translocation of food
3. Amphivasal (c) Fern
4. Xylem (d) Secondary growth
5. Phloem (e) Conduction of water

Answer:
1. (c) Fem
2. (d) Secondary growth
3. (a) Dracaena
4. (e) Conduction of water
5. (b) Translocation of food.

V. Answer in a Sentence

Question 1.
What is the collateral vascular bundle?
Answer:
When xylem lies towards the centre and phloem lies towards the periphery, it is called the collateral vascular bundle.

Question 2.
Where does the carbon that is used in photosynthesis come from?
Answer:
Carbondioxide present in atmosphere.

Question 3.
What is the common step in the aerobic and anaerobic pathway?
Answer:
Glycolysis is the common step in the aerobic and anaerobic pathway.

Question 4.
Name the phenomenon by which carbohydrates are oxidized to release ethyl alcohol.
Answer:
Fermentation (Anaerobic respiration)

VI. Short Answer Questions

Question 1.
Give an account on a vascular bundle of dicot stem.
Answer:
The vascular bundles of dicot stem are:

  • Conjoint: Xylem and phloem lie on the same radius.
  • Collateral: Xylem lies towards the centre and phloem lies towards the periphery.
  • Endarch: Protoxylem lies towards the centre and metaxylem lies towards the periphery.
  • Open: The cambium is present in between xylem and phloem.

The vascular bundles are arranged in the form of a ring around the pith.

Question 2.
Write a short note on mesophyll.
Answer:
The tissue present between the upper and lower epidermis in leaf is called mesophyll. It is differentiated into palisade and spongy parenchyma.

Question 3.
Draw and label the structure of oxysomes.
Answer:
Samacheer Kalvi 10th Science Solutions Chapter 12 Plant Anatomy and Plant Physiology 1

Question 4.
Name the three basic tissues system in flowering plants.
Answer:
The three basic tissue system in flowering plants are:

  1. Dermal or Epidermal tissue system
  2. Ground tissue system
  3. Vascular tissue system

Question 5.
What is photosynthesis and wherein a cell does it occur?
Answer:
Photosynthesis is a process in which the green plants use sunlight and the green pigment chlorophyll, to synthesize, nutrients from carbon dioxide from air and water. The photosynthesis occurs in green parts of the plant such as leaves, stems and floral buds.

Question 6.
What is respiratory quotient?
Answer:
The ratio of volume of carbon dioxide liberated and the volume of oxygen consumed, during respiration is called Respiratory Quotient (R.Q)
\(\mathrm{R} . \mathrm{Q} .=\frac{\text { Volume of } \mathrm{CO}_{2} \text { liberated }}{\text { Volume of } \mathrm{O}_{2} \mathrm{consumed}}\)

Question 7.
Why should the light dependent reaction occur before the light independent reaction?
Answer:
During light dependent reaction photosynthesis pigment absorb the light energy and convert it into chemical energy ATP and NADPH2.
During light independent CO2 is reduced into carbohydrates with the help of ATP and NADPH2 produced during light dependent reaction.

Question 8.
Write the reaction for photosynthesis.
Answer:
Samacheer Kalvi 10th Science Solutions Chapter 12 Plant Anatomy and Plant Physiology 2

VII. Long Answer Questions

Question 1.
Differentiate the following
(a) Monocot root and Dicot root
(b) Aerobic and Anaerobic respiration.
Answer:
(a) Monocot root and Dicot root

Dicot Root Monocot Root
1. The Xylem is Tetrarch 1. The Xylem is Polyarch.
2. The conjunctive tissue is made up of parenchyma cells. 2. The conjunctive tissue is made up of sclerenchyma cells.
3. The young root contains a path, but in the old root, pith is absent. 3. Pith cells are made of parenchyma cells with intercellular spaces and contain abundant starch grains.
4. Cambium is present during secondary growth. 4. Cambium is absent.
5. Secondary growth is present. 5. Secondary growth is absent.

(b) Aerobic and Anaerobic respiration

Aerobic Anaerobic
1. Occur in the presence of oxygen. 1. Occurs, when oxygen is absent.
2. Carbon dioxide, water and ATP are produced. 2. Lactic acid, Ethanol and ATP are produced.
3. It consists of 3 steps:

  • Glycolysis
  • Kreb’s cycle
  • Electron transport chain
3. It consists of 2 steps:

  • Glycolysis
  • Fermentation
    (Ethyl alcohol or Lactic acid are produced)

Question 2.
Describe and name three stages of cellular respiration that aerobic organisms use to obtain energy from glucose.
Answer:
The three stages of Aerobic respiration are:
(i) Glycolysis (Glucose splitting): It is the breakdown of one molecule of glucose into two molecules of pyruvic acid. Glycolysis takes place in the cytoplasm of the cell. It is the first step of both aerobic and anaerobic respiration.

(ii) Krebs Cycle: This cycle occurs in the mitochondria matrix. At the end of glycolysis, 2 molecules of pyruvic acid enter into mitochondria. The oxidation of pyruvic acid into CO2 and water takes place through this cycle. It is also called the Tricarboxylic Acid Cycle (TCA).

(iii) Electron Transport Chain: This is accomplished through a system of electron carrier complex called electron transport chain (ETC) located on the inner membrane of the mitochondria. NADH2 and FADH2 molecules formed during glycolysis and Krebs cycle are oxidised to NAD+ and FAD+ to release the energy via electrons. As they move, the electron release energy which is trapped by ADP to synthesis ATP. This is called oxidative phosphorylation. In this O2 gets reduced to water.

Question 3.
How does the light – dependent reaction differ from the light – independent reaction? What are the end products and reactants in each? Where does each reaction occur within the chloroplast?
Answer:
Light – dependent photosynthesis is called Hill reaction or Light reaction. The Light independent reactions are called Biosynthetic phase.

Light-dependent reaction Light independent reaction
1. It is called Hill reaction or Light reaction. 1. It is called Dark reaction or Biosynthetic pathway or the Calvin cycle.
2. The reaction is carried out in Thylakoid membranes (Grana) of the chloroplast. 2. This reaction is carried out in the stroma of the chloroplast.
3. Photosynthetic pigments absorb the light energy and convert it into chemical energy ATP and NADPH2. 3. CO2 is reduced into carbohydrates with the help of light generated ATP and NADPH2.
4. It is carried out in the presence of light. 4. It is carried out in the absence of light.

In the light – dependent reaction, the chlorophyll absorbs the light energy and convert it into chemical energy ATP and NADPH2. In the light – independent reaction, CO2 is reduced into carbohydrates with the help of light generated ATP and NADPH2. The light – dependent reaction is carried out in the Grana of the chloroplast. The Light independent reaction is carried out in the stroma of the chloroplast.

VIII. Higher Order Thinking Skills(HOTS) Questions

Question 1.
The reactions of photosynthesis make up a biochemical pathway.
(A) What are the reactants and products for both light and dark reactions?
(B) Explain how the biochemical pathway of photosynthesis recycles many of its own reactions and identify the recycled reactants.
Answer:
(a) Light reactions are carried out in the Thylakoid membranes (Grana) of the chloroplast which use water and energy from the Sun to produce NADPH, ATP and Oxygen. The Dark reaction is carried out in the stroma of the chloroplast. It uses NADPH, ATP and CO2 and produces NADP + ADP + P and high energy sugars.

(b) The Light reactions use light to synthesize ATP and NADPH. The dark reaction or Calvin cycle uses these reactants to produce sugar from additional CO2 molecule. This cycle then produces NAP + ADP + P; (inorganic phosphate) which is used in the light reactions, with water molecules to produce ATP and NADPH again.

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Question 2.
Where do the light-dependent reaction and the Calvin cycle occur in the chloroplast?
Answer:
Light-dependent reaction takes place in the presence of light energy in thylakoid membranes (grana) of the chloroplast Calvin cycle occurs at the stroma of the chloroplast.

Samacheer Kalvi 10th Science Plant Anatomy and Plant Physiology Additional Questions Solved

I. Choose the correct answer

Question 1.
The father of Plant Anatomy is _____.
(a) Melvin Calvin
(b) C.N.R. Rao
(c) Robin Hill
(d) Nehemiah Grew.
Answer:
(d) Nehemiah Grew.

Question 2.
The passage cells are found in endodermis of:
(a) dicot stem
(b) monocot stem
(c) dicot root
(d) dicot leaf
Answer:
(c) dicot root

Question 3.
The vascular bundle consists of _____.
(a) Xylem and Phloem
(b) Hypodermis and Endodermis
(c) Cortex and Pericycle
(d) Pith and Stele.
Answer:
(a) Xylem and Phloem

Question 4.
The vascular bundles are skull shaped in:
(a) dicot root
(b) monocot root
(c) dicot stem
(d) monocot stem
Answer:
(d) monocot stem

Question 5.
The protoxylem lacuna is present in the vascular bundles of:
(a) dicot root
(b) monocot root
(c) dicot stem
(d) monocot stem
Answer:
(d) monocot stem

II. Fill in the blanks

Question 1.
The epidermis has many minute pores called _____.
Answer:
Stomata.

Question 2.
Epiblema, the outermost layer of the root is called ____ or _____ layer.
Answer:
Rhizodermis or Piliferous.

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Question 3.
All the tissues, inner to Endodermis constitute _____.
Answer:
Stele.

Question 4.
______ provides mechanical support to plants.
Answer:
Sclerenchyma.

Question 5.
The barrel-shaped innermost layer of Dicot stem Endodermis is also called _____.
Answer:
Starch Sheath.

Question 6.
Each vascular bundle of monocot stem is surrounded by a few-layer of sclerenchyma cells called _____.
Answer:
Bundle Sheath

Question 7.
______ consists of sieve tubes and elements of companion cells.
Answer:
Phloem.

III. State whether the statements are true or false. Correct the false statement.

Question 1.
Pith is differentiated in monocot stems.
Answer:
False.
Correct Statement: Pith is not differentiated in monocot stems.

Question 2.
In monocot leaves, the mesophyll is not differentiated into palisade and spongy parenchyma.
Answer:
True.

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Question 3.
Dicot leaf is an isobilateral leaf.
Answer:
False.
Correct Statement: Monocot leaf is an isobilateral leaf.

Question 4.
The lateral roots of dicot plant originate from the stele.
Answer:
False.
Correct Statement: The lateral roots of dicot plant originate from the pericycle.

Question 5.
Cuticle and Stomata are absent in Epiblema of Dicot root.
Answer:
True.

IV. Match the following

Question 1.

1. Leukoplast (a) Photosystems
2. Accessory pigments (b) Inner mitochondrial membrane
3. Chlorophyll (c) Chlorophyll and Carotenoids
4. Cristae (d) Colourless plastids
5. Chl.a and Accessory pigments (e) Green pigment

Answer:
1. (d) Colourless plastids
2. (c) Chlorophyll and Carotenoids
3. (e) Green pigment
4. (b) Inner mitochondrial membrane
5. (a) Photosystems.

V. Short Answer Questions

Question 1.
Where does the break down of pyruvate to give carbondioxide, water and energy takes place?
Answer:
The break down of pyruvate to give carbondioxide. water and energy takes place in Mitochondria.

Question 2.
Draw the overview of Hill and Calvin Cycle.
Answer:
Samacheer Kalvi 10th Science Solutions Chapter 12 Plant Anatomy and Plant Physiology 3

Question 3.
Name the energy currency in the living organism. When and where it is produced?
Answer:
Adenosine Triphosphate (ATP) is called the energy currency in the living organism. It is produced in mitochondria during the process of respiration.

Question 4.
What are Bulliform cells?
Answer:
Some of the upper epidermal cells of monocot leaves are large and thin-walled. So they are called Bulliform cells.

Question 5.
Label the parts of the transverse section of Dicot Root.
Answer:
Samacheer Kalvi 10th Science Solutions Chapter 12 Plant Anatomy and Plant Physiology 4

Question 6.
Define conjuctive tissue.
Answer:
The tissue present between xylem and phloem is called conjuctive tissue. In monocot the conjuctive tissue is sclerenchymatous tissue and in dicot it is parenchymatous tissue.

Question 7.
List out the two important factors which affect photosynthesis.
Answer:

  1. Internal Factors:
    • Pigments
    • Leafage
    • Accumulation of carbohydrates
    • Hormones
  2. External Factors:
    • Light
    • Carbon dioxide
    • Temperature
    • Water
    • Mineral elements

Question 8.
Mention the components and functions of different Tissue Systems.
Answer:

Tissue System Components Functions
Dermal Tissue System Epidermis and Periderm (in older stems and roots)
  • Protection
  • Prevention of water loss
Ground Tissue System
  • Parenchyma tissue
  • Collenchyma tissue
  • Sclerenchyma tissue
  • Photosynthesis
  • Food storage
  • Regeneration
  • Support
  • Protection
Vascular Tissue System
  • Vascular tissues
  • Phloem tissue
  • Xylem tissue
  • Transport of water and minerals
  • Transport of food

VI. Long Answer Questions

Question 1.
(a) Label the parts of the Transverse section of a monocot stem.
(b) Mention the differences between Dicot and Monocot stem.
Answer:
(a)
Samacheer Kalvi 10th Science Solutions Chapter 12 Plant Anatomy and Plant Physiology 5
(b)

Tissues Dicot Stem Monocot stem
1. Hypodermis collenchymatous Sclerenchymatous
2. Ground tissue Differentiated into cortex, endodermis, pericycle and pith Undifferentiated
3. Vascular bundles
  • Less in number
  • Uniform in size
  • Arranged in a ring
  • Open
  • Bundle sheath absent
  • Numerous
  • Smaller near periphery, bigger in the centre
  • Scattered
  • Closed
  • Bundle sheath present
4. Secondary growth Present Mostly absent
5. Pith Present Absent
6. Medullary rays Present Absent

Question 2.
With a labelled diagram, explain the structure and function of mitochondria.
Answer:
Mitochondrial Membranes: It consists of two membranes called inner and outer membrane. Each membrane is 60 – 70 A° thick. The outer mitochondrial membrane is smooth and freely permeable to most small molecules. It contains enzymes, proteins and lipids. It has porin molecules (proteins) which form channels for passage of molecules through it.
Samacheer Kalvi 10th Science Solutions Chapter 12 Plant Anatomy and Plant Physiology 6
The inner mitochondrial membrane is semi – permeable membrane and regulates the passage of materials into and out of the mitochondria. It is rich in enzymes and carrier proteins. It consists of 80% of proteins and lipids.

Cristae: The inner mitochondrial membrane gives rise to finger-like projections called cristae. These cristae increase the inner surface area (fold in the inner membrane) of the mitochondria to hold a variety of enzymes.

Oxysomes: The inner mitochondrial membrane bear minute regularly spaced tennis racket shaped particles known as oxysomes (F1 particle). They involve in ATP synthesis.
Samacheer Kalvi 10th Science Solutions Chapter 12 Plant Anatomy and Plant Physiology 7
Mitochondrial matrix: It is a complex mixture of proteins and lipids. The matrix contains enzymes for Krebs cycle, mitochondrial ribosomes(70 S), tRNAs and mitochondrial DNA.

Question 3.
(a) Draw and label the ultrastructure of a chloroplast.
(b) Write the structure and function of the chloroplast.
Answer:
(a) Ultra Structure of a Chloroplast.
Samacheer Kalvi 10th Science Solutions Chapter 12 Plant Anatomy and Plant Physiology 8

(b) Structure and function of Chloroplast.
Chloroplasts are green plastids, containing the green pigment called Chlorophyll. It has the following structures:

  1. Envelope: It has outer and inner membranes, which are separated by intermembrane space.
  2. Stroma: Matrix present inside to the membrane is called stroma, which contains DNA, 70 S ribosomes and other molecules needed for protein synthesis.
  3. Thylakoids: It consists of thylakoid membrane that encloses thylakoid lumen. Thylakoid forms a stack of disc – like structures called granum.
  4. Grana: Some of the thylakoids are arranged in the form of discs, stacked one above the other called grana. These stacks are termed as grana, they are interconnected to each other by membranous lamellae called Fret channels.

Functions:

  • Photosynthesis
  • Storage of starch
  • Synthesis of fatty acids
  • Storage of lipids
  • Formation of chloroplasts.

VIII. Higher Order Thinking Skills (HOTS) Questions

Question 1.
What are non – green plants? Give an example.
Answer:
The plants, which do not have the green pigment, Chlorophyll and so they cannot prepare their food independently are called non-green plants. Eg. Orchids lack Chlorophyll.

Question 2.
Protophloem is the first formed phloem. If the protophloem surrounds by xylem, what kind of arrangement of phloem would you call it Give example.
Answer:
If the protophloem is surrounded by xylem the vascular bundle is said to be Concentric Amphivasal Vascular bundles. Eg: Dracaena

Question 3.
How does photosynthesis take place on plants that have, red, brown and yellow leaves, and are not green? How do they prepare food?
Answer:
These leaves have a much larger concentration of chromoplasts, which synthesize and store pigments such as orange carotenes, yellow xanthophylls and other red pigments. But these leaves, still possess chlorophyll ‘a’ and chlorophyll ‘b’ and can synthesize food. But these leaves have more chromoplasts and so the leaves do not appear green. Yet they can prepare food.

Question 4.
The cross-section of a plant material shown the following features on viewing under the microscope.
(a) Radially arranged Vascular bundles
(b) Xylem is exarch and polyarch
(c) Metaxylem is polygonal in shape.
Identify the slide.
Answer:
The given features are characters of dicot root

Samacheer Kalvi 10th Science Solutions Chapter 7 Atoms and Molecules

You can Download Samacheer Kalvi 10th Science Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 7 Atoms and Molecules

Samacheer Kalvi 10th Science Atoms and Molecules Textual Solved Problems

I. Calculation of molar mass:

Question 1.
Calculate the gram molar mass of the following.
(i) H2O
(ii) CO2
(iii) Ca3(PO4)2
Solution:
(i) H2O
Atomic masses of H = 1, O = 16
Gram molar mass of H2O = (1 × 2) + (16 × 1) = 2 + 16
Gram molar mass of H2O = 18 g.

(ii) CO2
Atomic masses of C = 12, O = 16
Gram molar mass of CO2 = (12 × 1) + (16 × 2) = 12 + 32
Gram molar mass of CO2 = 44 g.

(iii) Ca3(PO4)2
Atomic masses of Ca = 40, P = 30, O = 16.
Gram molar mass of Ca3(PO4)2 = (40 × 3) + [30 + (16 × 4)] × 2
= 120 + (94 × 2)
= 120 + 188
Gram molar mass of Ca3(PO4)2 = 308 g.

You can Download Samacheer Kalvi 10th Science Guide Pdf Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

II. Calculation based on number of moles from mass and volume:

Question 1.
Calculate the number of moles in 46 g of sodium.
Solution:
Number of moles = \(\frac{\text { Mass of the element }}{\text { Atomic mass of the element }}\)
Atomic mass of the element = \(\frac { 46 }{ 23 }\) = 2 moles of sodium

Question 2.
5.6 litre of Oxygen at S.T.P?
Solution:
Given volume of O2 at,
Number of moles = \(\frac{\text { S.T.P. }}{\text { Molar volume at } \mathrm{S} . \mathrm{T} \cdot \mathrm{P}}\)
Molar volume at S.T.P = \(\frac { 46 }{ 23 }\) = 2 moles
Number of moles of oxygen = \(\frac{5.6}{22.4}\) = 0.25 mole of oxygen

Question 3.
Calculate the number of moles of a sample that contains 12.046 × 1023 atoms of iron?
Solution:
Number of moles = \(\frac{\text { Number of atoms of iron }}{\text { Avogadro’s number }}\)
= 12.046 × 1023 / 6.023 × 1023
= 2 moles of iron.

III. Calculation of mass from a mole.

Question 1.
0.3 mole of aluminium (Atomic mass of Al = 27).
Solution:
Number of moles = \(\frac{\text { Mass of Al }}{\text { Atomic mass of Al }}\)
Mass = No. of moles × atomic mass
So, mass of Al = 0.3 × 27 = 8.1 g.

Question 2.
2.24 litre of SO2 gas at S.T.P?
Solution:
Molecular mass of SO2 = 32 + (16 × 2) = 32 + 32 = 64
Number of moles of SO2 = \(\frac{\text { Given volume of } \mathrm{SO}_{2} \text { at } \mathrm{S} . \mathrm{T.P}}{\text { Molar volume } \mathrm{SO}_{2} \text { at } \mathrm{S} . \mathrm{T.P}}\)
= \(\frac{2.24}{22.4}=0.1 \mathrm{mole}\)
Number of moles = \(\frac{\text { Mass }}{\text { Molecular mass }}\)
Mass = No. of moles × molecular mass
Mass = 0.1 × 64
Mass of SO2 = 6.4 g.

Question 3.
1.51 × 1023 molecules of water
Solution:
Molecular mass of H2O = 18
Number of moles = \(\frac{\text { Number of molecules of water }}{\text { Avogadro’s number }}\)
= 1.51 × 1023 / 6.023 × 1023 = 1 / 4 = 0.25 mole
Number of moles = \(\frac{\text { Mass }}{\text { Molecular mass }}\)
0.25 = mass / 18
Mass = 0.25 × 18
Mass = 4.5 g.

Question 4.
5 × 1023 molecules of glucose?
Solution:
Molecular mass of glucose = 180
Mass of glucose = \(\frac{\text { Molecular mass } \times \text { number of particles }}{\text { Avogadro’s number }}\)
= (180 × 5 × 1023) / 6.023 × 1023
= 149.43 g.

IV. Calculation based on the number of atoms/molecules.

Question 1.
Calculate the number of molecules in 11.2 litre of CO2 at S.T.P
Solution:
Number of moles of CO2 = \(\frac{\text { Volume at S.T.P }}{\text { Molar volume }}\)
= \(\frac { 11.2 }{ 22.4 }\)
= 0.5 mole.
Number of molecules of CO2 = Number of moles of CO2 × Avogadro’s number
= 0.5 × 6.023 × 1023
= 3.011 × 1023 molecules of CO2.

Question 2.
Calculate the number of atoms present in 1 gram of gold (Atomic mass of Au = 198).
Solution:
Number of atoms of Au = \(\frac{\text { Mass of Au } \times \text { Avogadro’s number }}{\text { Atomic mass of Au }}\)
Atomic mass of Au = \(\frac{1}{198} \times 6.023 \times 10^{23}\)
Number of atoms of Au = 3.042 × 1021 g.

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Question 3.
Calculate the number of molecules in 54 gm of H2O
Solution:
Number of molecules = \(\frac{(\text { Avogadro number } \times \text { Given mass })}{\text { Gram molecular mass }}\)
Number of molecules of water = 6.023 × 1023 × \(\frac { 54 }{ 18 }\)
= 18.069 × 1023 molecules.

Question 4.
Calculate the number of atoms of oxygen and carbon in 5 moles of CO2.
Solution:

  • 1 mole of CO2 contains 2 moles of oxygen.
  • 5 moles of CO2 contains 10 moles of oxygen
    Number of atoms of oxygen = number of moles of oxygen × Avogadro’s number
    = 10 × 6.023 × 1023 = 6.023 × 1024 atoms of Oxygen.
  • 1 mole of CO2 contains 1 mole of carbon
  • 5 moles of CO2 contains 5 moles of carbon
    No. of atoms of carbon = No.of moles of carbon × Avogadro’s number
    = 5 × 6.023 × 1023 = 3.011 × 1024 atoms of Carbon.

V. Calculation based on molar volume

Calculate the volume occupied by:
Question 1.
2.5 mole of CO2 at S.T.P.
Solution:
\(\begin{array}{l}{\text { Number of moles of } \mathrm{CO}_{2}=\frac{\text { Given volume at S.T.P }}{\text { Molar volume at S.T.P }}} \\ {\qquad 2.5 \text { mole of } \mathrm{CO}_{2}=\frac{\text { Volume of } \mathrm{CO}_{2} \text { at } \mathrm{S} . \mathrm{TP}}{22.4}}\end{array}\)
Volume of CO2 at S.T.P = 22.4 × 2.5 = 56 litres.

Question 2.
3.011 × 1023 of ammonia gas molecules?
Solution:
Number of moles = \(\frac{\text { Number of molecules }}{\text { Avogadro’s number }}\)
= 3.011 × 1023 / 6.023 × 1023
= 2 moles
Volume occupied by NH3 = number of moles × molar volume
= 2 × 22.4 = 44.8 litres at S.T.P.

Question 3.
14 g nitrogen gas?
Solution:
Number of moles = \(\frac { 14 }{ 28 }\) = 0.5 mole
Volume occupied by N2 at S.T.P = No. of moles × molar volume
= 0.5 × 22.4
= 11.2 litres.

VI. Calculation based on % composition.

Question 1.
Calculate % of S in H2SO4
Solution:
Molar mass of H2SO4 = (1 × 2) + (32 × 1) + (16 × 4)
= 2 + 32 + 64
= 98 g.
\(\begin{array}{l}{\% \text { of } \mathrm{S} \text { in } \mathrm{H}_{2} \mathrm{SO}_{4}=\frac{\text { Mass of sulphur }}{\text { Molar mass of } \mathrm{H}_{2} \mathrm{SO}_{4}} \times 100} \\ {\% \text { of } \mathrm{S} \text { in } \mathrm{H}_{2} \mathrm{SO}_{4}=\frac{32}{98} \times 100}\end{array}\)
= 32.65 %.

Samacheer Kalvi 10th Science Atoms and Molecules Textual Evaluation Solved

I. Choose the best answer.

Question 1.
Which of the following has the smallest mass?
(a) 6.023 × 1023 atoms of He
(b) 1 atom of He
(c) 2 g of He
(d) 1 mole atoms of He.
Answer:
(b) 1 atom of He
Hint:
(a) 6.023 × 1023 atoms of He = 1 mole
Mass of 1 mole of He = 4 g (or) 0.004 kg.

(b) Mass of 1 atom of He =?
Mass of 6.023 × 1023 atoms of He = 0.004 kg.
Mass of 1 atom of He = \(\frac{0.004}{6.023 \times 10^{23}}=6.6423 \times 10^{-27} \mathrm{kg}\)

(c) 2 g of He = Mass = 0.002 kg.

(d) 1 mole atoms of He = 4 g = 0.004 kg.
So (b) is the smallest mass as 6.6423 × 10-27 kg.

Question 2.
Which of the following is a triatomic molecule?
(a) Glucose
(b) Helium
(c) Carbon dioxide
(d) Hydrogen
Answer:
(c) Carbon dioxide

Question 3.
The volume occupied by 4.4 g of CO2 at S.T.P _____.
(a) 22.4 litre
(b) 2.24 litre
(c) 0.24 litre
(d) 0.1 litre.
Answer:
(b) 2.24 litre
Hint:
Molar volume of CO2 = 22.4 litre.
The volume occupied by 1 mole.
i.e. 44 g (molar mass) of CO2.
44 g of CO2 occupied 22.4 litre of volume.
4.4 g of CO2 will occupy \(\frac{22.4}{44}\) × 4.4 = \(\frac{22.4}{10}\) = 2.24 litre.
So, answer (b) is correct.

Question 4.
Mass of 1 mole of Nitrogen atom is:
(a) 28 amu
(b) 14 amu
(c) 28 g
(d) 14 g
Answer:
(c) 28 g

SamacheerKalvi.Guru

Question 5.
Which of the following represents 1 amu?
(a) Mass of a C – 12 atom
(b) Mass of a hydrogen atom
(c) \(\frac { 1 }{ 2 }\)th of the mass of a C – 12 atom
(d) Mass of O – 16 atom
Answer:
(c) 1 / 12th of the mass of a C – 12 atom
Hint: By definition 1 amu is defined as precisely 1 / 12th the mass of an atom of carbon – 12.
So, answer (c) is correct.

Question 6.
Which of the following statement is incorrect?
(a) One gram of C – 12 contains Avogadro’s number of atoms.
(b) One mole of oxygen gas contains Avogadro’s number of molecules.
(c) One mole of hydrogen gas contains Avogadro’s number of atoms.
(d) One mole of electrons stands for 6.023 × 1023 electrons.
Answer:
(a) One gram of C – 12 contains Avogadro’s number of atoms.

Question 7.
The volume occupied by 1 mole of a diatomic gas at S.T.P is _____.
(a) 11.2 litre
(b) 5.6 litre
(c) 22.4 litre
(d) 44.8 litre.
Answer:
(c) 22.4 litre
Hint: By definition 1 mole of any gas at S.T.P occupies molar volume i.e. 22.4 litres.
So (c) is the correct answer.

Question 8.
In the nucleus of \(_{20} \mathrm{Ca}^{40}\), there are _____.
(a) 20 protons and 40 neutrons
(b) 20 protons and 20 neutrons
(c) 20 protons and 40 electrons
(d) 40 protons and 20 electrons.
Answer:
(b) 20 protons and 20 neutrons
Hint:
\(_{20} \mathrm{Ca}^{40}\)
20 = Atomic number = Number of protons (or) Number of electrons
40 = Mass number = Number of protons + Number of neutrons
\(_{20} \mathrm{Ca}^{40}\) contains 20 protons, 20 electrons and 20 neutrons.
So the answer (b) is correct.

Question 9.
The gram molecular mass of oxygen molecule is:
(a) 16 g
(b) 18 g
(c) 32 g
(d) 17 g
Answer:
(b) 18 g

Question 10.
1 mole of any substance contains ______ molecules.
(a) 6.023 × 1023
(b) 6.023 × 10-23
(c) 3.0115 × 1023
(d) 12.046 × 1023.
Answer:
(a) 6.023 × 1023
Hint:
Avogadro’s law states that 1 mole of any substance contains 6.023 × 1023 molecules.
So the answer (a) is correct.

II. Fill in the blanks

Question 1.
Atoms of different elements having ______ mass number, but ______ atomic numbers are called isobars.
Answer:
Same, different.

Question 2.
Atoms of different elements having the same number of _____ are called isotones.
Answer:
Neutrons.

Question 3.
Atoms of one element can be transmuted into atoms of other elements by _____.
Answer:
Artificial transmutation.

Question 4.
The sum of the numbers of protons and neutrons of an atom is called its _____.
Answer:
Mass number.

SamacheerKalvi.Guru

Question 5.
Relative atomic mass is otherwise known as _____.
Answer:
Standard atomic weight.

Question 6.
The average atomic mass of hydrogen is _____ amu.
Answer:
1.008 amu.

Question 7.
If a molecule is made of similar kind of atoms, then it is called ______ atomic molecule.
Answer:
Homo.

Question 8.
The number of atoms present in a molecule is called its _____.
Answer:
Atomicity.

Question 9.
One mole of any gas occupies _____ ml at S.T.P.
Answer:
22400.

Question 10.
Atomicity of phosphorous is _____.
Answer:
4.

III. Match the following.

Question 1.

a. 8 g of O2 i. 4 moles
b. 4 g of H2 ii. 0.25 moles
c. 52 g of He iii. 2 moles
d. 112 g of N2 iv. 0.5 moles
e. 35.5 g of Cl2 v. 13 moles

Answer:
a – ii, b – iii, c – v, d – i, e – iv.

IV. True or False: (If false give the correct statement)

Question 1.
Two elements sometimes can form more than one compound.
Answer:
True.

Question 2.
Noble gases are Diatomic
Answer:
False.
Correct Statement: Noble gases are monoatomic

Question 3.
The gram atomic mass of an element has no unit?
Answer:
False.
Correct Statement: The gram atomic mass of an element is expressed in the unit grams.

Question 4.
1 mole of Gold and Silver contain the same number of atoms?
Answer:
True

Question 5.
The molar mass of CO2 is 42 g?
Answer:
False.
Correct Statement: The molar mass of CO2 is (12 + 32) = 44 g.

V. Assertion and Reason:

Answer the following questions using the data given below:
(i) A and R are correct, R explains the A.
(ii) A is correct, R is wrong.
(iii) A is wrong, R is correct.
(iv) A and R are correct, R doesn’t explain A.

Question 1.
Assertion: Atomic mass of aluminium is 27
Reason: An atom of aluminium is 27 times heavier than 1 / 12th of the mass of the C – 12 atoms.
Answer:
(i) A and R are correct, R explains the A.

Question 2.
Assertion: The Relative Molecular Mass of Chlorine is 35.5 a.m.u.
Reason: The natural abundance of Chlorine isotopes are not equal.
Answer:
(i) A and R are correct, R explains the A.

VI. Short Answer Questions

Question 1.
Define Relative atomic mass.
Answer:
Relative atomic mass of an element is the ratio between the average mass of its isotopes to 1 / 12th part of the mass of a carbon – 12 atom. It is denoted as Ar.
\(\text { Relative atomic mass }=\frac{\text { Average mass of the isotopes of the element }}{1 / 12^{\text {th }} \text { of the mass of one Carbon- } 12 \text { atom }}\).

Question 2.
Write the different types of isotopes of oxygen and its percentage abundance.
Answer:
Isotopes of oxygen:
Samacheer Kalvi 10th Science Solutions Chapter 7 Atoms and Molecules 1
The atomic mass of oxygen = (15.9949 × 0.99757) + (16.9991 × 0.00038) + (17.9992 × 0.00205) = 15.999 amu.

Question 3.
Define Atomicity.
Answer:
The number of atoms present in the molecule is called atomicity.

Question 4.
Give any two examples for heteroatomic molecules.
Answer:
Heterodiatomic molecules.
e.g.,

  • HCl
  • NaCl.

Question 5.
What is Molar volume of a gas?
Answer:
The volume occupied by one mole of any gas at STP is called molar volume. Its value is equal to 22.4 litre or 22400 ml or 22400 cm³ or 2.24 × 10-2 m³.

Question 6.
Find the percentage of nitrogen in ammonia.
Answer:
Ammonia – NH3 = Molar mass = 14 + 3 = 17
Mass % of Nitrogen = \(\frac{14}{17} \times 100\) = 82.35%.

VII. Long Answer Questions.

Question 1.
Calculate the number of water molecule present in one drop of water which weighs 0.18 g.
Answer:
One mole of water weighs 18 g.
18 g of water contains 6.023 × 1023 water molecules.
∴ 0.18 g of water contains
= \(\frac{6.023×10^{23}}{18}\) × 0.18
= 6.023 × 1021 water molecules

Question 2.
N2 + 3H2 → 2NH3 (The atomic mass of nitrogen is 14, and that of hydrogen is 1)

  • 1 mole of nitrogen (____ g) + ____.
  • 3 moles of hydrogen (____ g) → ____.
  • 2 moles of ammonia (____ g).

Answer:
N2 + 3H2 → 2NH3

  • 1 mole of N2 = 28 g
  • 3 moles of H2 = 6 g
  • 2 moles of NH3 = 34 g
    ⇒ 1 mole of nitrogen (28 g) + 3 moles of hydrogen (6 g) → 2 moles of Ammonia (34 g).

Question 3.
Calculate the number of moles in:
(i) 27 g of Al
(ii) 1.51 × 1023 molecules of NH4Cl
Answer:
(i) 27 g of Al
Number of moles = \(\frac{\text { Mass }}{\text { Atomic mass }}\)
Number of moles in 27 g of Al = \(\frac{27}{27}\) = 1 mole.

(ii) 1.51 × 1023 molecules of NH4Cl
Number of moles = \(\frac{\text { Number of molecules }}{\text { Avogadro’s number }}=\frac{1.51 \times 10^{23}}{6.023 \times 10^{23}}=0.25 \text { moles. }\).

Question 4.
Give the salient features of ‘Modern atomic theory’.
Answer:
(i) An atom is no longer indivisible.
(ii) Atoms of the same element may have different atomic mass.
Eg: isotopes 17Cl35, 17Cl37.
(ii) Atoms of different elements may have same atomic masses.
Eg: Isobars 18Ar40, 20Ca40.
(iv) Atoms of one element can be transmuted into atoms of other elements. An atom is no longer indestructible.
(v) Atoms may not always combine in a simple whole number ratio.
Eg: Glucose C6H12O6 C : H : O = 6 : 12 : 6 or 1 : 2 : 1.
(vi) Atom is the smallest particle that takes part in a chemical reaction.
(vii) The mass of an atom can be converted into energy (E = mc²).

Question 5.
Derive the relationship between Relative molecular mass and Vapour density.
Answer:
(i) The Relative Molecular Mass of a gas or vapour is the ratio between the mass of one molecule of the gas or vapour to mass of one atom of Hydrogen.

(ii) Vapour density is the ratio of the mass of a certain volume of a gas or vapour, to the mass of an equal volume of hydrogen, measured under the same conditions of temperature and pressure.
Vapour Density (V.D.) = \(\frac{\text { Mass of a given volume of gas or vapour at S.T.P }}{\text { Mass of same volume of hydrogen }}\).

(iii) According to Avogadro’s law, equal volumes of all gases contain equal number of molecules. Thus, let the number of molecules in one volume = n, then

(iv) V.D at STP = \(\frac{\text { Mass of “n’molecules of a gas or vapour at S.T.P }}{\text { Mass of ‘n’molecules of hydrogen }}\)
Cancelling ‘n’ which is common, you get
V.D = \(\frac{\text { Mass of } 1 \text { molecule of a gas or vapour at S.T.P. }}{\text { Mass of } 1 \text { molecules of hydrogen }}\).

(v) Since hydrogen is diatomic
\(\mathrm{V} . \mathrm{D} .=\frac{\text { Mass of } 1 \text { molecule of a gas or vapour at S.T.P. }}{\text { Mass of } 2 \text { atoms of hydrogen }}\).

(vi) By comparing the definition of relative molecular mass and vapour density we can write as follows.
\(\mathrm{V.D.}=\frac{\text { Mass of } 1 \text { molecule of a gas or vapour at S.T.P. }}{2 \times \text { Mass of } 1 \text { atom of hydrogen }}\)
Relative molecular mass (hydrogen scale) \(=\frac{\text { Mass of } 1 \text { molecule of a gas or vapour at STP. }}{\text { Mass of } 1 \text { atom of hydrogen }}\).

(vii) By substituting the relative molecular mass value in vapour density definition, we get
Vapour density (V.D.) = Relative molecular mass / 2
⇒ 2 × vapour density = Relative molecular mass of a gas.

VIII. HOT Questions

Question 1.

Calcium carbonate is decomposed on heating in the following reaction CaCO3 → CaO + CO2
(i) How many moles of Calcium carbonate are involved in this reaction?
Answer:
One mole

(ii) Calculate the gram molecular mass of calcium carbonate involved in this reaction.
Answer:
Gram molecular mass of CaCO3
= 40 + 12 + 3(16)
= 100 g

(iii) How many moles of CO2 are there in this equation?
Answer:
One mole.

SamacheerKalvi.Guru

IX. Solve the following problems.

Question 1.
How many grams are there in the following?
Answer:
Formula = No. of moles (n) × (Gram molecular mass)

(i) 2 moles of hydrogen molecule, H2
Answer:
Mass of 2 moles of H2 molecule
= 2 × 2 = 4 g

(ii) 3 moles of chlorine molecule, Cl2
Answer:
Gram molecular mass of 3 moles of Cl2
= 3 × 71 = 213 g

(iii) 5 moles of sulphur molecule, S2
Answer:
Gram molecular mass of 5 moles of S2
= 5 × 8(32)
= 5 × 256 = 1280 g

(iv) 4 moles of phosphorous molecule, P4
Answer:
Gram molecular mass of 4 moles of P2
= 4 × 4(31)
= 4 × 124 = 496 g

Question 2.
Calculate the % of each element in calcium carbonate. (Atomic mass: C – 12, O – 16, Ca – 40)
Solution:
Calcium carbonate: CaCO3
Molar mass of CaCO3 = 40 + 12 + (16 × 3) = 100 g
% of Calcium \(=\frac{40}{100} \times 100=40 \%\)
% of Carbon \(=\frac{12}{100} \times 100=12 \%\)
% of Oxygen \(=\frac{48}{100} \times 100=48 \%\).

Question 3.
Calculate the % of oxygen in Al2(SO4)3. (Atomic mass: Al – 12, O – 16, S – 32)
Solution:
Aluminium Sulphate – Al2(SO4)3
Molar mass of Aluminium Sulphate = (27 × 2) + (32 × 3) + (16 × 12) = 54 + 96 + 192 = 342 g
% of Oxygen \(=\frac{192}{342} \times 100=56.14 \%\).

Question 4.
Calculate the % relative abundance of B – 10 and B – 11, if its average atomic mass is 10.804 amu.
Solution:
The average atomic mass of Boron = 10.804 amu.
% relative abundance of B – 10 = ?
% relative abundance of B – 11 = ?
Let the fraction of relative abundance of B – 10 = x
Let the fraction of relative abundance of B – 11 = y
x + y = 1
y = 1 –  x
Relative abundance = x (10) + (1 – x) (11) = 10.804 amu
⇒ 10x + 11 – 11x = 10.804 amu
⇒ 11 – x = 10.804 amu
⇒ -x = 10.804 – 11
⇒ -x = -0.196
⇒ x = 0.196
x = % abundance of B – 10 = 0.196 × 100 = 19.6 %
y = % abundance of B – 11 = 100 – 19.6 = 80.4 %
Percentage abundance of B – 10 = 19.6 %
Percentage abundance of B – 11 = 80.4 %.

Activities

Question 1.
Complete the following table by filling the appropriate values / terms
Samacheer Kalvi 10th Science Solutions Chapter 7 Atoms and Molecules 2
Solution:
Samacheer Kalvi 10th Science Solutions Chapter 7 Atoms and Molecules 3

Question 2.
Classify the following molecules based on their atomicity and fill in the table:
Fluorine (F2), Carbon dioxide (CO2), Phosphorous (P4), Sulphur (S8), Ammonia (NH3), Hydrogen iodide (HI), Sulphuric Acid (H2SO4), Methane (CH4), Glucose (C6H12O6), Carbon monoxide (CO)
Samacheer Kalvi 10th Science Solutions Chapter 7 Atoms and Molecules 4
Solution:
Samacheer Kalvi 10th Science Solutions Chapter 7 Atoms and Molecules 5

Question 3.
Under same conditions of temperature and pressure if you collect 3 litres of O2, 5 litres of Cl2 and 6 litres of H2,

  1. Which has the highest number of molecules?
  2. Which has the lowest number of molecules?

Solution:
Number of moles of O2 \(=\frac{\text { Volume at S.T.P }}{\text { Molar volume }}\) \(=\frac{3}{22.4}\) = 0.1339 moles
Number of molecules = Number of moles × Avogadro number
= 0.1339 × 6.023 × 1023
= 0.8064 × 1023
= 8.064 × 1022 O2 molecules.
Number of moles of Cl2 = \(\frac{5}{22.4}\) = 0.2232 moles
Number of molecules = 0.2232 × 6.023 × 1023 = 1.344 × 1023 molecules.
Number of moles of H2 = \(\frac{6}{22.4}\) = 0.2678 moles
Number of molecules = 0.2678 × 6.023 × 1023 = 1.6129 × 1023 molecules.

  1. 6 litres of H2 has the highest number of molecules.
  2. 3 litres of O2 has the lowest number of molecules.

Samacheer Kalvi 10th Science Atoms and Molecules Additional Question Solved

I. Choose the best answer.

Question 1.
Which of the following pair indicates isotopes?
(a) \(_{17} \mathrm{Cl}^{35},_{17} \mathrm{Cl}^{37}\)
(b) \(_{18} \mathrm{Ar}^{40},_{20} \mathrm{Ca}^{40}\)
(c) \(_{6} \mathrm{C}^{13},_{7} \mathrm{N}^{14}\)
(d) \(_{33} \mathrm{As}^{77},_{34} \mathrm{Se}^{78}\).
Answer:
(a) \(_{17} \mathrm{Cl}^{35},_{17} \mathrm{Cl}^{37}\)

Question 2.
The mass of a proton is equal to:
(a) 1 amu
(b) \(\frac{1}{12^{th}}\) of the mass of a C – 12 atom
(c) zero
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

Question 3.
The sum of the number of protons and neutrons of an atom is called its _____.
(a) nucleus
(b) atomic number
(c) mass number
(d) relative atomic mass.
Answer:
(c) mass number

Question 4.
Total number of electrons present in 1.7 g of NH3 is:
(a) 6.023 × 1023
(b) 6.023 × 1024
(c) 6.023 × 1022
(d) 6.023 × 1025
Answer:
(a) 6.023 × 1023

Question 5.
An isotope of hydrogen without neutrons is _____.
(a) Deuterium \(_{1} \mathrm{H}^{2}\)
(b) Protium \(_{1} \mathrm{H}^{1}\)
(c) Tritium \(_{1} \mathrm{T}^{3}\)
(d) Heavy hydrogen \(_{1} \mathrm{D}^{2}\).
Answer:
(b) Protium \(_{1} \mathrm{H}^{1}\)

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Question 6.
The isotope tritium contains 1 proton and neutron in the nucleus.
(a) 1
(b) 2
(c) 3
(d) none
Answer:
(b) 2

Question 7.
Which one of the following element is used as the standard for measuring the relative atomic mass of an element in now a days?
(a) \(_{1} \mathrm{H}^{2}\)
(b) \(6^{\mathrm{O}^{12}}\)
(c) C – 12
(d) C – 14.
Answer:
(c) C – 12

Question 8.
The atom with no neutrons in the nucleus is:
(a) He
(b) Deuterium
(c) Tritium
(d) Protium
Answer:
(d) Protium

Question 9.
The average atomic mass of carbon is _____.
(a) 12 amu
(b) 12.84 amu
(c) 24.011 amu
(d) 12.011 amu.
Answer:
(d) 12.011 amu.

Question 10.
Which one of the following is the most abundant element in both the Earth’s crust and in the human body?
(a) Carbon
(b) Silicon
(c) Oxygen
(d) Hydrogen.
Answer:
(c) Oxygen

Question 11.
Gram molecular mass of H2SO4 is:
(a) 49 g
(b) 54 g
(c) 98 g
(d) 100 g
Answer:
(c) 98 g

Question 12.
Boron – 10 and Boron – 11 are called _____.
(а) isotopes
(b) isobars
(c) isotones
(d) isomers.
Answer:
(c) isotopes

Question 13.
Which of the following are found in the elementary state in nature?
(a) Hydrogen chloride
(b) Carbon dioxide
(c) Noble gases
(d) Oxygen.
Answer:
(c) Noble gases

Question 14.
Ammonia gas is formed by the following reaction
N2(g) + 3H2(g) → 2NH3(g)
The volume of H2 required to form 6 dm3 of NH3 is:
(a) 9 dm³
(b) 10 dm³
(c) 4 dm³
(d) 2 dm³
Answer:
(a) 9 dm³

SamacheerKalvi.Guru

Question 15.
Which one of the following is a hetero diatomic molecule?
(a) O2
(6) N2
(c) HI
(d) CH4.
Answer:
(c) HI

Question 16.
Which one of the following is a hetero triatomic molecule?
(a) H2O
(b) BCl3
(c) CH4
(d) PCl5.
Answer:
(a) H2O

Question 17.
1 gm atom of nitrogen represents:
(a) 6.023 × 102 N2 molecules
(b) 22.4 litre of N2 at STP
(c) 11.2 L of N2 at STP
(d) 28 g of nitrogen
Answer:
(c) 11.2 L of N2 at STP

Question 18.
Find out the hetero diatomic molecule?
(a) Hydrogen
(b) Hydrogen chloride
(c) Methane
(d) Ammonia.
Answer:
(b) Hydrogen chloride

Question 19.
Which one of the following is an example of a polyatomic molecule?
(a) Sulphur
(b) Gold
(c) Sodium
(d) Helium.
Answer:
(a) Sulphur

Question 20.
The gram molar mass of CO2 is:
(a) 44 g
(b) 100 g
(c) 4.4 g
(d) 22 g
Answer:
(a) 44 g

Question 21.
Which one of the following is an example of a polyatomic molecule?
(a) Fluorine
(b) Glucose
(c) Oxygen
(d) Sodium.
Answer:
(b) Glucose (C6H12O6)

Question 22.
The gram molecular mass of water is _____.
(a) 18 amu
(b) 18 g
(c) 18 u
(d) 18.
Answer:
(b) 18 g

Question 23.
The value of Avogadro’s number is _____.
(a) 6.023 × 10-23
(b) 6.023 × 1023
(c) 22.4
(d) 22400.
Answer:
(b) 6.023 × 1023

Question 24.
The value of molar volume is _____.
(a) 22.4 ml
(b) 22.4 litres
(c) 22400 litres
(d) 2.24 litres.
Answer:
(b) 22.4 litres

Question 25.
Which one of the following represent Avogadro’s law?
(a) V ∝ \(\frac{1}{n}\)
(b) V ∝ n
(c) V ∝ \(\frac{1}{n^{2}}\)
(d) V2 ∝ \(\frac{1}{n}\).
Answer:
(b) V ∝ n

SamacheerKalvi.Guru

Question 26.
Which of the following has the highest number of molecules?
(a) 1 litre of N2
(b) 2 litres of oxygen
(c) 5 litres of Cl2
(d) 6 litres of Hydrogen.
Answer:
(d) 6 litres of Hydrogen.

Question 27.
Which one of the following has the lowest number of molecules?
(a) 1 litre of N2
(b) 2 litres of H2
(c) 3 litres of O2
(d) 4 litres of Cl2
Answer:
(a) 1 litre of N2

Question 28.
2 × Vapour density is equal to _____.
(a) atomic mass
(b) valency
(c) relative molecular mass
(d) atomic number.
Answer:
(c) relative molecular mass

Question 29.
The value of gram molar mass of CO2 is _____.
(a) 44 amu
(b) 44 g
(c) 44
(d) 44 kg.
Answer:
(b) 44 g
Hint: Molar mass = 12 + (16 × 2) = 44 g.

Question 30.
The number of moles of a sample that contain 36 g of water is _____.
(a) 1 mole
(b) 0.5 mole
(c) 4 moles
(d) 2 moles.
Answer:
(d) 2 moles
Hint: 18 g of water = 1 mole
36 g of water = \(\frac{1}{18} \times 36\) = 2 moles

Question 31.
Which of the following has the largest number of particles?
(a) 8 g of CH4
(b) 4.4 g of CO2
(c) 34.2 g of C12H22O11
(d) 2 g of H2.
Answer:
(d) 2 g of H2.
Hint. 2 g = Molar mass = 1 mole = 6.023 × 1023 particles.
Others are less.

Question 32.
The number of molecules in 16.0 g of oxygen is _____.
(a) 6.023 × 1023
(b) 6.023 × 10-23
(c) 3.01 × 10-23
(d) 3.011 × 1023
Answer:
(d) 3.011 × 1023
Hint: 32 g of oxygen contain 6.023 × 1023 molecules.
16 g of oxygen will contain
\(\frac{6.023 \times 10^{23}}{32} \times 16=3.011 \times 10^{23}\)

Question 33.
The percentage of hydrogen in H2O is _____.
(a) 8.88
(b) 11.2
(c) 20.60
(d) 80.0.
Answer:
(b) 11.2
Hint: 1 mole of H2O has 2.016 g of H2
Percentage of H2 = \(\frac{2.016}{18}\) × 100 = 11.2

Question 34.
Which of the following contains the largest number of molecules?
(a) 0.2 mole of H2
(b) 8.0 g of H2
(c) 17 g of H2O
(d) 6.0 g of CO2
Answer:
(b) 8.0 g of H2
Hint: No. of moles = \(\frac{8}{2}\) = 4 moles.
No. of molecules = mole × Avogadro number = 4 × 6.023 × 1023 = 2.409 × 1024

Question 35.
One gram of which of the following contains the largest number of oxygen atoms?
(a) O
(b) O2
(c) O3
(d) All contain the same
Answer:
(c) O3

SamacheerKalvi.Guru

Question 36.
The percentage by weight of O2 in CaSO4. (O = 16, Ca = 40, S = 32) is _____.
(a) 64 %
(b) 28.2 %
(c) 47.05 %
(d) 16.2 %.
Answer:
(c) 47.05 %
Hint: % by weight of O2 = \(\frac{64}{136}\) × 100 = 47.05 %.

Question 37.
One mole of a gas occupies a volume of 22.4 L. This is derived from _____.
(a) Berzilliu’s hypothesis
(b) Gay – Lussac’s law
(c) Avogadro’s law
(d) Dalton’s law.
Answer:
(c) Avogadro’s law

Question 38.
Volume of gas at STP is 1.12 × 10-7 cc. Calculate the number of molecules in it.
(a) 3.011 × 1020
(b) 3.011 × 1012
(c) 3.011 × 1023
(d) 3.011 × 1024
Answer:
(b) 3.011 × 1012
Hint. 2.24 × 10-3 c molecules 6.023 × 1023 molecules
1.12 × 10-7 cc contains = \(\frac{6.023 \times 10^{23}}{22400} \times 1.12 \times 10^{-7}\)
= 3.011 × 1012.

Question 39.
The number of molecules of CO2 present in 44 g of CO2 is _____.
(a) 6.023 × 1023
(b) 3.011 × 1023
(c) 12 × 1023
(d) 3 × 1010.
Answer:
(a) 6.023 × 1023 (Avogadro number).

Question 40.
The volume occupied by 4.4 g of CO2 at S.T.P is _____.
(a) 22.4 L
(b) 2.24 L
(c) 0.224 L
(d) 0.1 L.
Answer:
(b) 2.24 L
Hint. 44 g of CO2 at S.T.P occupies 22.4 L
4.4 g of CO2 at S.T.P will occupy \(\frac{22.4}{44}\) × 4.4 = 2.24 L.

Question 41.
How many molecules at present in one gram of hydrogen?
(a) 6.023 × 1023
(b) 3.011 × 1023
(c) 2.5 × 1023
(d) 1.5 × 1023
Answer:
(b) 3.011 × 1023
Hint: H2 = Molar mass = 2 g
2 g of H2 contains 6.023 × 1023 molecules
∴ 1 g of H2 will contain = \(\frac{6.023 \times 10^{23}}{2} \times 1\)
= 3.011 × 1023 molecules.

Question 42.
Atoms which have the same number of protons but different number of neutrons are called _____.
(a) isotopes
(b) isomers
(c) allotropes
(d) isotones.
Answer:
(a) isotopes

Question 43.
Number of atoms which a molecule to sulphur contains is _____.
(a) 3
(b) 8
(c) 4
(d) 2.
Answer:
(b) 8 (S8)

Question 44.
An example of a triatomic molecule is _____.
(a) Ozone
(b) Nitrogen
(c) Hydrogen
(d) Ammonia.
Answer:
(a) Ozone

Question 45.
The atomic mass of sodium is 23. The number of moles in 46 g of sodium is _____.
(a) 0.5
(b) 2
(c) 1
(d) 0.25.
Answer:
(b) 2
Hint:
No. of moles = \(\frac{\text { Mass }}{\text { Atomic mass }}=\frac{46}{23}\) = 2.

Question 46.
The number of atoms in a molecule of the elementary substance is called _____.
(a) Atomic number
(b) Avogadro number
(c) Atomic mass
(d) Atomicity.
Answer:
(d) Atomicity.

Question 47.
Avogadro number represents the number of atoms in _____.
(a) 12 g of C – 12
(b) 4.4 g of CO2
(c) 320 g of Sulphur
(d) 1 g of C – 12
Answer:
(a) 12 g of C – 12

Question 48.
The number of moles in 5 grams of Calcium is _____.
(a) 0.5 mole
(b) 0.125 mole
(c) 1.25 mole
(d) 12.5 mole.
Answer:
(a) 0.125 mole
Hint:
No. of moles = \(\frac{\text { Mass }}{\text { Atomic mass }}\)
\(=\frac{5}{40}=\frac{1}{8}\) = 0.125 mole.

Question 49.
One mole of H2O corresponds to _____.
(a) 22.4 litre at 1 atm and 250°C
(b) 6.023 × 1023 atoms of hydrogen and 6.023 × 1023 atoms of oxygen
(c) 18 g
(d) 1 g.
Answer:
(c) 18 g
Hint: One mole = Molar mass = 2 + 16 = 18 g.

Question 50.
Which one of the following has the maximum number of atoms?
(a) 18 g of H2O
(b) 18 g of O2
(c) 18 g of CO2
(d) 18 g of CH4.
Answer:
(d) 18 g of CH4.
Hint:
Samacheer Kalvi 10th Science Solutions Chapter 7 Atoms and Molecules 6
Samacheer Kalvi 10th Science Solutions Chapter 7 Atoms and Molecules 7

Question 51.
The atomicity of K2Cr2O7 is _____.
(a) 9
(b) 11
(c) 10
(d) 12.
Answer:
(b) 11

Question 52.
All noble gases are _____ molecules.
(a) diatomic
(b) triatomic
(c) mono atomic
(d) poly atomic.
Answer:
(c) mono atomic

Question 53.
The total number of atoms represented by the compound CuSO4 . 5H2O is ____.
(a) 27
(b) 21
(c) 5
(d) 8.
Answer:
(b) 21

Question 54.
Which one of the following represents the mass of 0.5 moles of water molecules?
(a) 18 g
(b) 1.8 g
(c) 9 g
(d) 4.5 g.
Answer:
(c) 9 g
\(\text { Mole }=\frac{\text { Mass }}{\text { Molecular mass }}\)
Mass = Mole × Molecular mass = 0.5 × 18 = 9 g.

Question 55.
The atomic mass of Calcium is 40. Calculate the number of moles in 16 g of Calcium.
(a) 0.4 mole
(b) 4 moles
(c) 640 moles
(d) \(\frac { 1 }{ 4 }\) mole.
Answer:
(a) 0.4 mole
Hint:
\(\text { Mole }=\frac{\text { Mass }}{\text { Atomic mass }}=\frac{16}{40}=\frac{8{}}{20}\) \(=\frac{4}{10}=0.4 \mathrm{mole}\).

Question 56.
If the atomic mass of sodium is 23 amu, then the mass of 3.011 × 1023 sodium atoms is _____.
(a) 11.5 kg
(b) 23 g
(c) 0.5 mole
(d) 11.5 g.
Answer:
(d) 11.5 g.
Hint: Mass of 6.023 × 1023 sodium atoms = 23 amu = 23 g.
∴ Mass of 3.011 × 1023 sodium atoms
\(=\frac{23}{6.023 \times 10^{23}} \times 3.011 \times 10^{23}=11.5 \mathrm{g}\).

Question 57.
Which of the following will have maximum mass?
(а) 0.1 mole of NH2
(b) 1022 atoms of carbon
(c) 1022 molecules of CO2
(d) 1 g of Fe
Answer:
(a) 0.1 mole of NH3
Hint:
(a) 0.1 mole of NH3 has 6.023 × 1023 atoms.
Mass of 1 mole of NH3 = 17 g
Mass of 0.1 mole of NH3 = 1.7 g.

(b) Mass of 1022 atoms of carbon
6.023 × 1023 c atoms mass = 12 g
1022 atoms of C has the mass
\(=\frac{12}{6.023 \times 10^{23}} \times 1022=2.036 \times 10^{-20} \mathrm{g}\).

(c) Mass of 1022 molecules of CO2
CO2 = molar mass = 44 g
6.023 × 1023 CO2 molecules has the mass = 44 g
∴ 1022 CO2 molecules has the mass 44
\(=\frac{44}{6.023 \times 10^{23}} \times 1022=7.466 \times 10^{-20} \mathrm{g}\).

(d) 1 g of Fe
∴ (a) 1.7 g of NH3 has the highest mass.

Question 58.
Which of the following correctly represents 360 g of water?
(i) 2 moles of water
(ii) 20 moles of water
(iii) 6.023 × 1023 molecules of water
(iv) 1.2044 × 1025 molecules of water
(a) (i) only
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii) and (iv).
Answer:
(d) (ii) and (iv).
Hint: (i) 2 moles of water
Mass of 1 mole of water = 18 g
Mass of 2 moles of water = 18 × 2 = 36 g.

(ii) 20 moles of water
Mass of 1 mole of water = 18 g
Mass of 20 moles of water = 18 × 20 = 360 g.

(iii) 6.023 × 1023 molecules of water = 1 mole = 18 g.

(iv) 1.2044 × 1025 molecules of water
6.23 × 1023 molecules of water = 1 mole
∴ 1.2044 × 1025 molecules
\(=\frac{1}{6.023 \times 10^{23}} \times 1.2044 \times 10^{25}\)
= 20 moles.
∴ Mass of 20 moles = 20 × 18 = 360 g.
So (d) is correct.

Question 59.
Which of the following contains maximum number of molecules?
(a) 1 g of CO2
(b) 1 g of N2
(c) 1 g of H2
(d) 1 g of CH4
Answer:
(b) 1 g of H2
Hint:
(a) 1 g of CO2
No. of moles = \(\frac{\text { Mass }}{\text { Atomic mass }}\) (or) \(\frac{\text { Mass }}{\text { Molecular mass }}\)
No. of moles of 1 g of CO2 = \(\frac{1}{44}\)
No. of molecules = \(\frac{1}{44}\) × 6.023 × 1023
= 1.368 × 1022 molecules of CO2.

(b) 1 g of N2
No. of molecules = \(\frac{1}{28}\) × 6.023 × 1023
= 2.151 × 1022 molecules of N2.

(c) 1 g of H2
No. of molecules = \(\frac{6.023 \times 10^{23} \times 1}{2}\)
= 3.011 × 1023 molecules of H2

(d) 1 g of CH4
No. of molecules = \(\frac{6.023 \times 10^{23} \times 1}{16}\)
= 3.764 × 1022 molecules of CH4
So (c) is the correct answer.

Question 60.
Which of the following pair is an example of isotopes?
\(\begin{array}{l}{\text { (a) } 21 \mathrm{Sc}^{45} \text { and }_{23} \mathrm{V}^{50}} \\ {\text { (b) }_{22} \mathrm{Ti}^{48} \text { and }_{22} \mathrm{Ti}^{50}} \\ {\text { (c) }_{22} \mathrm{Ti}^{50} \text { and }_{23} \mathrm{V}^{50}} \\ {\text { (d) }_{21} \mathrm{Sc}^{45} \text { and }_{22} \mathrm{Ti}^{50}}\end{array}\)
Answer:
(b) \(\text { (b) }_{22} \mathrm{Ti}^{48} \text { and }_{22} \mathrm{Ti}^{50}\).

SamacheerKalvi.Guru

II. Fill in the blanks.

Question 1.
Amedeo Avogadro put forward a hypothesis based on the relation between the number of _____ and the _____ of gases.
Answer:
Molecules, volume.

Question 2.
The molar volume of a gas at STP is _____ and the value of Avogadro Number is _____.
Answer:.
22.4.litres, 6.023 x 1023.

Question 3.
Nitrogen and oxygen are _____ molecules whereas Helium and Neon are ____ molecules.
Answer:
Diatomic, monoatomic

Question 4.

  1. _____ are the building blocks of matter.
  2. ______ is a triatomic molecule.

Answer:

  1. Atoms and molecules
  2. Ozone

Question 5.
NH3, H2O are _____ molecules whereas N2, O2 are _____ molecules.
Answer:
Heteroatomic, Homoatomic

Question 6.
____ and ____ are polyatomic molecules.
Answer:
Phosphorous (P4), Sulphur (S8)

Question 7.

  1. Atoms of the same element with same atomic number but a different mass number are called _____.
  2. Atoms of different elements with the same number of neutrons are called _____.

Answer:

  1. Isotopes
  2. Isotones

Question 8.
Atomicity of Nitrogen is _____ whereas the atomicity of Helium is _____.
Answer:
2, 1.

Question 9.
Atoms of the same element with same atomic number but having different mass number are called _____.
Answer:
Isotopes.

Question 10.
Atoms of different elements with the same atomic mass but a different atomic number are called _____.
Answer:
Isobars.

Question 11.
Atoms of different elements having the same number of neutrons but a different atomic number and different mass number are called _____.
Answer:
Isotones.

Question 12.
_____ is the smallest particle that takes part in the chemical reaction.
Answer:
Atom.

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Question 13.
Anything that has mass and occupies space is called _____.
Answer:
Matter.

Question 14.
Protons and neutrons have considerable mass, but _____ don’t have considerable mass.
Answer:
Atoms.

Question 15.
_____ is one-twelfth of the mass of C – 12 atom, an isotope of carbon which contains _____ protons and ____ neutrons.
Answer:
The atomic mass unit, 6, 6.

Question 16.
_____ are the building blocks of matter.
Answer:
Atoms.

Question 17.
The stable isotope of _____ is used as the standard for measuring the relative atomic mass of an element.
Answer:
Carbon C – 12.

Question 18.
Modem methods of determination of atomic mass by _____ use C – 12 as standard.
Answer:
Mass Spectrometry.

Question 19.
The relative atomic mass of sulphur is _____.
Answer:
32.

Question 20.
The average atomic mass of carbon is ______.
Answer:
12.011 amu.

Question 21.
The average atomic mass of an element becomes fractional due to the presence of ______.
Answer:
Isotopes.

Question 22.
_____ is the most abundant element in both the Earth’s crust and the human body.
Answer:
Oxygen.

Question 23.
Except for _____ atoms of most of the elements are found in the combined form with itself or atoms of other elements.
Answer:
Noble gases.

Question 24.
A molecule is a combination of two or more atoms held together by _____.
Answer:
Chemical bonds.

Question 25.
If the molecule is made of similar kind of atoms, it is called ______.
Answer:
Homo atomic molecule.

SamacheerKalvi.Guru

Question 26.
The molecule that consists of atoms of different elements is called _____.
Answer:
Hetero atomic molecule.

Question 27.
The number of _____ present in the molecule is called its atomicity.
Answer:
atoms.

Question 28.
The atomicity of ozone is _____.
Answer:
3.

Question 29.
The atomicity of hydrogen chloride is _____.
Answer:
2.

Question 30.
Water is a _____ molecule.
Answer:
Hetero triatomic.

Question 31.
One mole of an element contains ______ atoms and it is equal to its gram atomic mass.
Answer:
6.023 × 1023

Question 32.
One mole of any gas occupies ______ or _____ at S.T.P.
Answer:
22.4 litre, 22400 ml.

Question 33.
The _____ is useful to determine the empirical formula and molecular formula.
Answer:
Percentage composition.

Question 34.
The percentage composition of elements is useful to determine _____ and _____.
Answer:
Empirical formula, molecular formula.

Question 35.
Avogadro’s law is in agreement with ______.
Answer:
Dalton’s atomic theory.

Question 36.
_____ determines the relation between molecular mass and vapour density.
Answer:
Avogadro’s law.

Question 37.
Relative molecular mass is equal to _____.
Answer:
2 × Vapour density.

Question 38.
Atomicity of sulphur is _____.
Answer:
8.

Question 39.
The metals Cu, Ag, Au are _____ elements.
Answer:
Monoatomic.

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Question 40.
The atomicity of H2SO4 is ______.
Answer:
7.

Question 41.
Atomicity of an element is equal to _____.
Answer:
\(\frac{\text { Molecular mass }}{\text { Atomic mass }}\)

III. Match the following.

Question 1.

i. Monoatomic molecule (a) Ozone
ii. Diatomic molecule (b) Phosphorous
iii. Triatomic molecule (c) Helium
iv. Polyatomic molecule (d) Oxygen

Answer:
i – c, ii – d, iii – a, iv – b.

Question 2.

i. 22.4 litres (a) Avogadro Number
ii. 6.023 × 1023 (b) Molar volume
iii. 2 × vapour density (c) 1 mole
iv. Mass / Atomic mass (d) Molecular mass

Answer:
i – b, ii – a, iii – d, iv – c.

Question 3.

i. \(_{17} \mathrm{Cl}^{35},_{17} \mathrm{Cl}^{37}\) (a) Isotones
ii. \(_{6} \mathrm{Cl}^{13},_{7} \mathrm{N}^{14}\) (b) Isobars
iii. \(_{18} \mathrm{Ar}^{40},_{20} \mathrm{Ca}^{40}\) (c) E = mc2
iv. Einstein’s equation (d) Isotopes

Answer:
i – d, ii – a, iii – b, iv – c.

Question 4.

i. H2O (a) 180 g
ii. NH3 (b) 44 g
iii. CO2 (c) 17 g
iv. C6H12O6 (d) 18 g

Answer:
i – d, ii – c, iii – b, iv – a.

Question 5.

i. NH3, CH4 (a) Polyatomic molecule
ii. O2, N2 (b) Monoatomic molecule
iii. He, Ne (c) Heteroatomic molecule
iv. Sulphur (d) Diatomic molecule

Answer:
i – c, ii – d, iii – b, iv – a.

Question 6.

i. F2 (a) Polyatomic molecule
ii. O3 (b) Monoatomic molecule
iii. P4 (c) Diatomic molecule
iv. He (d) Triatomic molecule

Answer:
i – c, ii – d, iii – a, iv – b.

Question 7.

i. H2 (a) Hetero diatomic molecule
ii. HCl (b) Monoatomic molecule
iii. H2O (c) Homo diatomic molecule
iv. Ne (d) Hetero triatomic molecule

Answer:
i – c, ii – a, iii – d, iv – b.

Question 8.

i. Isotopes (a) S8, P4
ii. Isobars (b) \(_{6} \mathrm{C}^{13},_{7} \mathrm{N}^{14}\)
iii. Isotones (c) \(_{1} \mathrm{H}^{1},_{1} \mathrm{H}^{2},_{1} \mathrm{H}^{3}\)
iv. Polyatomic molecule (d) \(_{18} \mathrm{Ar}^{40},_{20} \mathrm{Ca}^{40}\)

Answer:
i – c, ii – d, iii – b, iv – a.

Question 9.

i. H2O (a) 16
ii. CO2 (b) 18
iii. C6H12O6 (c) 44
iv. CH4 (d) 180

Answer:
i – b, ii – c, iii – d, iv – a

Question 10.

i. 22 g of CO2 (a) 2 moles
ii. 18 g of H2O (b) 4 moles
iii. 360 g of Glucose (c) 0.5 mole
iv. 64 g of CH4 (d) 1 mole

Answer:
i – c, ii – d, iii – a, iv – b.

IV. State whether true or false. If false, give the correct statement.

Question 1.
Isotopes are the atoms of the same element may not be similar in all respects.
Answer:
True.

Question 2.
Isobars are the atoms of the different elements with the same atomic number and different mass numbers.
Answer:
False.
Correct statement: Isobars are the atoms of the different elements with the same mass number but a different atomic number.

Question 3.
Isotones are the atoms of different elements with the same number of neutrons.
Answer:
True.

Question 4.
The number of molecules present in one mole of an element is called atomicity of an element.
Answer:
False.
Correct statement: The number of atoms present in one molecule of an element is called the atomicity of an element.

SamacheerKalvi.Guru

Question 5.
Avogadro’s hypothesis is used in the deduction of atomicity of elementary gases.
Answer:
True.

Question 6.
The volume of a gas at a given temperature and pressure is proportional to the number of particles.
Answer:
True.

Question 7.
The value of Gram molar volume at STP is 11.2 litres.
Answer:
False.
Correct statement: The value of Gram molar volume at STP is 22.4 litres.

Question 8.
The atomicity of nitrogen, oxygen and hydrogen is two.
Answer:
True.

Question 9.
Atoms and molecules are the building blocks of matter.
Answer:
True.

Question 10.
The atoms of certain elements such as hydrogen, oxygen and nitrogen have an independent existence.
Answer:
False.
Correct statement: The atoms of certain elements such as hydrogen, oxygen and nitrogen do not have an independent existence.

Question 11.
A molecule is the simplest structural unit of an element or compound which contains one or more atoms.
Answer:
True.

Question 12.
Phosphorous and sulphur are monoatomic molecules.
Answer:
False.
Correct statement: Phosphorous and sulphur are polyatomic molecules.

Question 13.
H2O, NH3, CH4 are examples of homoatomic molecules.
Answer:
False.
Correct statement: H2O, NH3, CH4 are examples of heteroatomic molecules.

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Question 14.
An atom of one element can be transmuted into an atom of other element is known as artificial transmutation.
Answer:
True.

Question 15.
The molecule is the smallest particle that takes part in a chemical reaction.
Answer:
False.
Correct statement: Atom is the smallest particle that takes part in a chemical reaction.

Question 16.
The sum of the number of protons and neutrons of an atom is called Atomic number.
Answer:
False.
Correct statement: The sum of the number of protons and neutrons of an atom is called mass number.

Question 17.
The stable isotope of carbon (C – 12) with atomic mass 12 is used as the standard for measuring the relative atomic mass of an element.
Answer:
True.

Question 18.
The gram atomic mass of oxygen is 16 g.
Answer:
True.

Question 19.
Silicon is the most abundant element in the Earth’s crust.
Answer:
False.
Correct statement: Oxygen is the most abundant element in the Earth’s crust.

Question 20.
Except for noble gases, atoms of most of the elements are found in the combined form.
Answer:
True.

Question 21.
The number of atoms present in the molecule is called the Avogadro number.
Answer:
False.
Correct statement: The number of atoms present in the molecule is called its Atomicity.

Question 22.
O2, N2, H2, Cl2, Br2, F2, I2 are hetero diatomic molecules.
Answer:
False.
Correct statement: O2, N2, H2, Cl2, Br2, F2, I2 are homo diatomic molecules.

Question 23.
Water is an example of Hetero triatomic molecule.
Answer:
True.

Question 24.
One molecule of an element contains 6.023 × 1023 atoms and it is equal to its gram atomic mass.
Answer:
True.

Question 25.
An equal volume of all gases under similar conditions of temperature and pressure contain a different number of molecules.
Answer:
False.
Correct statement: Equal volume of all gases under similar conditions of temperature and pressure contain the same number of molecules.

Question 26.
The mathematical representation of Avogadro’slawisV/n=Constant(or)Vccn(or) V = Constant × n.
Answer:
True.

Question 27.
The molecular formula of gases can be derived using Avogadro’s law.
Answer:
True.

Question 28.
The number of moles of a sample that contains 12.046 x 1023 atoms of iron is 2.
Answer:
True.

Question 29.
The volume occupied by 14 g of Nitrogen gas is 22.4 litres.
Answer:
False.
Correct statement: The volume occupied by 14 g of Nitrogen gas is 11.2 litres.

Question 30.
Avogadro’s law determines the relation between molecular mass and absolute density.
Answer:
False.
Correct statement: Avogadro’s law determines the relation between molecular mass and vapour density.

V. Assertion and Reason

Question 1.
Assertion (A): C12H22O11 is not a simple ratio.
Reason (R): The ratio of atoms in a molecule may be fixed and integral but may not be simple.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 2.
Assertion (A): \(_{6} \mathrm{C}^{13}\) and \(_{7} \mathrm{N}^{4}\) are called Isotones.
Reason (R): Isotones are the atoms of the different elements with different atomic number but the same mass number.
(a) Both (A) and (R) are correct
(b)Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(c) (A) is correct but (R) is wrong

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Question 3.
Assertion (A): Nitrogen, oxygen and hydrogen are diatomic molecules.
Reason (R): N2, O2, H2 contain two atoms in one molecule and so they are a diatomic molecule.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 4.
Assertion (A): Atoms and molecules are the building blocks of matter.
Reason (R): Atom is the ultimate particle of an element which may or may not have an independent existence.
(a) Both (A) and (R) are wrong]
(b) (A) is correct but (R) does not explain (A)
(c) Both (A) and (R) are correct
(d) (A) is wrong but (R) is correct.
Answer:
(c) Both (A) and (R) are correct

Question 5.
Assertion (A): Hydrogen, Oxygen and Ozone are called homoatomic molecules.
Reason (R): Homoatomic molecules are made up of atoms of the same element.
(a) Both (A) and (R) are wrong
(b) (A) is correct but (R) is wrong
(c) (A) is wrong but (R) is correct
(d) Both (A) and (R) are correct.
Answer:
(d) Both (A) and (R) are correct.

Question 6.
Assertion (A): Water, Ammonia (H2O, NH3) are heteroatomic molecules.
Reason (R): Most of the elementary gases and compounds consist of atoms of the same element.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(c) (A) is correct but (R) is wrong

Question 7.
Assertion (A): 18 g water contains Avogadro number (6.023 × 1023) of particles.
Reason (R): 18 g of water is the molecular mass (or) 1 mole of water. One mole is defined as the amount of the substance which contains 6.023 × 1023 number of particles.
(a) (A) is correct and (R) explains (A)
(b) (A) is correct but (R) is wrong
(c) (A) is wrong but (R) is correct
(d) Both (A) and (R) are wrong
Answer:
(a) (A) is correct and (R) explains (A)

Question 8.
Assertion (A): Atoms of the same element may not be similar in all respects.
Reason (R): Atoms of the same element have the same atomic number but a different number of neutrons.
(a) Both (A) and (R) are correct
(b) (A) is correct but (R) is wrong
(c) Both (A) and (R) are wrong
(d) (A) is wrong but (R) is correct.
Answer:
(b) (A) is correct but (R)is wrong

Question 9.
Assertion (A): The atomicity of ozone is three.
Reason (R): 1 molecule of ozone contains 3 atoms of oxygen.
(a) Both (A) and (R) are correct
(b) Both (A) and(R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 10.
Assertion (A): \(_{1} \mathrm{H}^{1}, \quad_{1} \mathrm{H}^{2},_{1} \mathrm{H}^{3}\) are the isotopes of hydrogen.
Reason (R): The atoms of the same element with the same mass number but different at numbers are called isotopes.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(c) (A) is correct but (R) is wrong

Assertion (A) & Reason (R):
(i) (A) and (R) are correct. (R) explain (A)
(ii) (A) is correct (R) is wrong
(iii) (A) is wrong (R) is correct
(iv) (A) and (R) are correct. (R) does not explain (A).

Question 11.
Assertion (A): An atom is no longer indivisible.
Reason (R): The subatomic particles protons, electrons and neutrons were discovered.
Answer:
(i) (A) and (R) are correct. (R) explain (A)

Question 12.
Assertion (A): \(_{18} \mathrm{Ar}^{40}\) and \(_{20} \mathrm{Ca}^{40}\) are isobars.
Reason (R): They have the same atomic mass but a different atomic number.
Answer:
(i) A) and (R) are correct; (R) explain (A)

Question 13.
Assertion (A): \(_{17} \mathrm{Cl}^{35}\) and \(_{17} \mathrm{Cl}^{37}\) are isotones.
Reason (R): Atoms of the same element have the same atomic number but a different mass number.
Answer:
(iii) (A) is wrong (R) is correct

Question 14.
Assertion (A): NH3, H2O, HCl are heteroatomic molecules.
Reason (R): The molecule that consists of atoms of different elements is called heteroatomic molecules.
Answer:
(i) (A) and (R) are correct. (R) explain (A)

Question 15.
Assertion (A): \(_{6} \mathrm{C}^{13}\) and \(_{7} \mathrm{N}^{14}\) are called isotones.
Reason (R): Atoms of different elements having the same number of neutrons, but a different atomic number and different mass number are called isotones.
Answer:
(i) (A) and (R) are correct. (R) explain (A)

VI. Short Answer Questions.

Question 1.
What are isotopes? Give an example.
Answer:
Atoms of the same element that have same atomic number but different mass number are called isotopes.
e.g., \(_{1} \mathrm{H}^{1},_{1} \mathrm{H}^{2},_{1} \mathrm{H}^{3}\).

Question 2.
State Avogadro Hypothesis.
Answer:
Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.

SamacheerKalvi.Guru

Question 3.
What are isotones? Give an example.
Answer:
Atoms of different elements having the same number of neutrons but a different atomic number and different mass numbers are called isotones.
e.g., \(_{6} \mathrm{C}^{13},_{7} \mathrm{N}^{14}\).

Question 4.
Define Mole.
Answer:
Mole is defined as the amount of substance that contains as many specified elementary particles as the number of atoms in 12 g of C-12 isotope.
It is also defined as the amount of substance which contains Avogadro number (6.023 × 1023) of particles.

Question 5.
Define

  1. Atomic number
  2. Mass number

Answer:

  1. The atomic number of an element is the number of protons or number of neutrons and electrons present in it.
  2. The mass number is the sum of the number of protons and neutrons in an atom.

Question 6.
How many grams are there in
(i) 5 moles of H2O
Answer:
5 moles of H2O = 5 × 18 = 90 g

(ii) 1 mole of Glucose (C6H12O6)
Answer:
1 mole of Glucose (C6H12O6) = 180 g

Question 7.
Define molecule.
Answer:
A molecule is a combination of two or more atoms held together by the strong chemical force of attraction, i.e. Chemical bonds.

Question 8.
What is homo atomic molecule? Give two examples.
Answer:
If the molecule is made of similar kind of atoms, then it is called homoatomic molecule. e.g. H2, Cl2

Question 9.
What is a heteroatomic molecule? Give two examples.
Answer:
The molecule that consists of atoms of different elements is called a heteroatomic molecule. e.g. HCl, H2O

Question 10.
Consider the following and classify them on the basis of their atomicity.
H2, CCl4, O3, BF3, HCl, HNO3, C12H22O11, NO, Cl2, He, Au, P4

  • Monoatomic molecule – He, Au
  • Homo diatomic molecule – H2, Cl2
  • Homo triatomic molecule – O3
  • Homo polyatomic molecule – P4
  • Hetero diatomic molecule – HCl, NO
  • Hetero polyatomic molecule – CCl4, BF3, HNO3, C12H22O11.

Question 11.
Define Relative molecular mass.
Answer:
The Relative molecular mass of a molecule is the ratio between the mass of one molecule of the substance to 1 / 12th mass of an atom of Carbon – 12 isotope.

Question 12.
Define Mole.
Answer:
The mole is the amount of the substance that contains as many elementary entities as there are atoms in exactly 12 g of the carbon – 12 isotope.

Question 13.
Define the Avogadro number.
Answer:
The actual number of atoms in 12 g of carbon – 12 is called the Avogadro number.
It is equal to 6.023 × 1023 (NA).

Question 14.
What is meant by percentage composition? What is its use?
Answer:
The percentage composition of a compound represents the mass of each element present in 100 g of the compound. It is useful to determine the empirical formula and molecular formula.

Question 15.
State Avogadro hypothesis (or) Avogadro’s Law.
Answer:
The Avogadro’s law states that “equal volume of all gases under similar conditions of temperature and pressure contain the equal number of molecules”.
[V ∝ n].

Question 16.
What are the applications of Avogadro’s Law?
Answer:

  • It explains Gay – Lussac’s law.
  • It helps in the determination of atomicity of gases.
  • The molecular formula of gases can be derived using Avogadro’s law.
  • It determines the relation between molecular mass and vapour density.
  • It helps to determine the gram molar volume of all gases, (i.e, 22.4 litres at S.T.P).

Question 17.
How is Average atomic mass calculated?
Answer:
The average atomic mass of an element is calculated by adding the masses of its isotopes, each multiplied by their natural abundance on the Earth.

Question 18.
Define Vapour density.
Answer:
The vapour density is defined as the ratio between the masses of equal volumes of a gas (or vapour) and hydrogen under the same condition.

SamacheerKalvi.Guru

Question 19.
Write the relationship between

  1. Atomicity and Molecular mass
  2. Molecular mass and Vapour density.

Answer:

  1. Atomicity = \(\frac{\text { Molecular mass }}{\text { Atomic mass }}\)
  2. Molecular mass = 2 × Vapour density

Question 20.
Distinguish between isotopes and isobars.
Answer:

Isotopes Isobars
The atoms of the same element with same atomic number (Z) but different mass number (A) are called isotopes.
e.g. \(_{17} \mathrm{Cl}^{35},_{17} \mathrm{Cl}^{37}\)
The atoms of the different element with the same mass number (A) but different atomic number (Z) are called isobars.
e.g. \(_{18} \mathrm{Ar}^{40},_{20} \mathrm{Ca}^{40}\)

Question 21.
What are the types of molecules? Give an example for each type?
Answer:
Molecules are of two types:

  • Homoatomic molecule: The molecules which are made up of atoms of the same element are called Homoatomic molecule, e.g., N2, O2, H2
  • Heteroatomic molecule: The molecules which are made up of atoms of different elements are called Heteroatomic molecule, e.g., NH3, H2O, CH4

VII. HOT Questions.

Question 1.
Calculate the mass of CO2 which contains the same number of molecules as are contained in 40 g of SO2.
Answer:
Gram molecular mass of SO2 = 32 + 2(16)
= 64 g
No. of moles of SO2 = \(\frac{GivenMass}{Mol.Mass}\)
= \(\frac{40}{64}\) = 0.625 moles
∵ Equal moles contains equal number of molecules.
Mass of CO2 which contains the same number of molecules,
= 0.625 × mol. mass of CO2
= 0.625 × 44
= 27.5 g

Question 2.
A flask P contains 0.5 moles of oxygen gas. Another flask Q contains 0.4 moles of ozone gas. Which of the two flasks contains greater number of oxygen atoms?
Answer:
1 molecule of oxygen (O2) = 2 atoms of oxygen
1 molecule of ozone (O3) = 3 atoms of oxygen
In flask P:
1 mole of oxygen gas = 6.022 × 1023 molecules
0.5 mole of oxygen gas = 6.022 × 1023 × 0.5 molecules
= 6.022 × 1023 × 0.5 × 2 atoms
= 6.022 × 1023 atoms.

In flask Q:
1 mole of ozone gas = 6.022 × 1023 molecules
0.4 mole of ozone gas = 6.022 × 1023 × 0.4 molecules
= 6.022 × 1023 × 0.4 × 3 atoms
= 7.23 × 1022 atoms
Flask Q has a greater number of oxygen atoms as compared to the flask P.

Question 3.
Chlorophyll, the green pigment of plants responsible for photosynthesis contain 2.68% of Mg by weight. Calculate the number of magnesium atoms in 20 g of chlorophyll.
Answer:
The weight % of Mg as 2.68
i.e.,100 g of chlorophyll contains 2.68 g of Mg
∴ 2 g of chlorophyll will contain Mg
\(\frac{2.68}{100}\) × 20
= 0.5369
1 mole of Mg = 24 g = 6.023 × 1023 atoms
∴ 0.0536 g of Mg will have = \(\frac{6.023×10^{23}}{24}\) × 0.536
= 0.1345 × 1023 atoms of Mg = 1.345 × 1022
Number of Magnesium atoms present in 20 g of chlorophyll is 1.345 × 1022

Question 4.
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 C – atoms = 6 moles
  2. 1 mole of C2H6 contains 6 moles of hydrogen atoms
    3 moles of C2H6 will contain H-atoms = 18 moles
  3. 1 mole of C2H6 contains Avogadro’s number. i.e., 6.023 × 1023 molecules.
    3 moles of C2H6 will contain ethane molecules = 3 × 6.023 × 1023 = 18.06 × 1023 molecules.

Question 5.
If ten volumes of dihydrogen gas react with five volumes of dioxygen gas, how many volumes of water vapour could be produced?
Answer:
H2 and O2 react according to the equation
H2 (g) + O2 (g) → 2H2O (g)
Thus, 2 volumes of H2 react with 1 volume of O2 to produce 2 volumes of water vapour.
Hence, 10 volumes of H2 will react completely with 5 volumes of O2 to produce 10 volumes of water vapour.

VIII. Long Answer Questions.

Question 1.
What are the differences between atoms and molecules?
Answer:

Atom Molecule
An atom is the smallest particle of an element A molecule is the smallest particle of an element or compound.
Atom does not exist in the free state except in a noble gas The molecule exists in the free state
Except some of the noble gas, other atoms are highly reactive Molecules are less reactive
Atom does not have a chemical bond Atoms in a molecule are held by chemical bonds
Example: Na Example: N2

Question 2.
Write the applications of Avogadro’s Law.
Answer:
(i) It explains Gay-Lussac’s law.
(ii) It helps in the determination of atomicity of gases.
(iii) Molecular formula of the gases can be derived.
(iv) It determines the relation between molecular mass and vapour density.
(v) It helps to determine gram molar volume of all gases

SamacheerKalvi.Guru

Question 3.
State and explain the applications of Avogadro’s law.
(OR)
Give any two applications of Avogadro’s law.
(OR)
Write any three applications of Avogadro’s law.
Answer:
Avogadro’s law: Equal volumes of all gases under the same conditions of temperature and pressure contain an equal number of molecules.
Applications of Avogadro’s law:

  • It is used to determine the atomicity of gases.
  • It is helpful in determining the molecular formula of gaseous compounds.
  • It establishes the relationship between the vapour density and molecular mass of a gas.
  • It gives the value of the molar volume of gases at STP. Molar volume of a gas at STP = 22.4 litres.
  • It explains Gaylussac’s law effectively.

Question 4.
Explain the classification of molecules based on atomicity.
Answer:
In accordance with the number of atoms present in the molecules, they are classified as monoatomic, diatomic, triatomic and polyatomic molecules showing that they contain one, two, three or more than 3 atoms respectively.

Atomicity Number of atoms per molecule Example
Monoatomic molecule 1 Helium (He), Neon (Ne) metals (Fe, Cu)
Diatomic molecule 2 Hydrogen (H2), Chlorine (Cl2)
Triatomic molecule 3 Ozone (O3)
Polyatomic molecule >3 Phosphorous (P4), Sulphur (S8)

Question 5.
A compound made up of two elements A and B has A = 70%, B = 30%. Their relative number of moles in the compound are 1.25 and 1.88. Calculate.
(a) Atomic masses of the elements A and B.
(b) The molecular formula of the compound, if its molecular mass is found to be 160.
Answer:

Elements Relative no. of moles Simplest molar ratio Simplest whole no. molar ratio
A 1.25 \(\frac{1.25}{1.25}=1\) 2
B 1.88 \(\frac{1.88}{1.25}=1.5\) 3

(a) Atomic mass of A = \(\frac{70}{1.25}\) = 56
Atomic mass of B = \(\frac{30}{1.88}\) = 16

(b) The molecular mass of the compound = 160
The molecular formula of the compound = Fe2O3

IX. Solve the following problems.

Question 1.
Calculate the gram molar mass of the following.
(a) NaOH
(b) C12H22O11
(c) H3PO4
(Atomic mass of Na – 23, O -16, H – 1, C – 12, P – 31)
Answer:
(a) NaOH (Sodium hydroxide)
GMM = 23 + 16 + 1
= 40 g
Gram molar mass of NaOH = 40 g

(b) C12H22O11 (Sucrose)
GMM = 12 × 12 + 22 × 1 + 11(16)
= 342 g
Gram molar mass of sucrose = 342 g

(c) H3 PO4 (Phosphoric acid)
GMM = 3(1) + 1(31)+ 4(16) = 98 g
Gram molar mass of Phosphoric acid = 98 g.

Question 2.
Calculate the percentage composition of oxygen and hydrogen by taking the example of H2O
Solution:
Mass % of an element = \(\frac{\text { Mass of that element in the compound }}{\text { Molar mass of the compound }} \times 100\)
Now, Molar mass of H2O = 2(1) + 16 = 18 g
Mass % of Hydrogen = \(\frac{2}{18} \times 100\) = 11.11 %
Mass % of Oxygen = \(\frac{16}{18} \times 100\) = 88.89 %.

Question 3.
What is the mass of 1 atom of Gold? (At. mass of Au = 197)
Answer:
The mass of 6.023 × 1023 atoms of Gold = 197 g
∴ The mass of 1 atom of gold = \(\frac{197}{6.023×10^{23}}\) × 1
= 3.27 × 10-22 g

Question 4.
Find the gram molecular mass of the following from the data given:
(i) H2O
(ii) CO2
(iii) NaOH
(iv) NO2
(v) H2SO4

Element Symbol Atomic No. Atomic Mass
Hydrogen H 1 1
Carbon C 6 12
Oxygen O 8 16
Nitrogen N 7 14
Sodium Na 11 23
Sulphur S 16 32

Solution:
(i) H2O
Atomic mass of 2(H) = 2 × 1 = 2
Atomic mass of 1(O) = 1 × 16 = 16
Molecular mass of H2O = 2 + 16 = 18

(ii) CO2
Atomic mass of 1(C) = 1 × 12 = 12
Atomic mass of 2(O) = 2 × 16 = 32
Molecular mass of CO2 = 12 + 32 = 44 g

(iii) NaOH
Atomic mass of 1(Na) = 1 × 23 = 23
Atomic mass of 1(O) = 1 × 16 = 16
Atomic mass of 1(H) 1 × 1 = 1
Molecular mass of NaOH = 23 + 16 + 1 = 40 g

(iv) NO2
Atomic mass of 1(N) = 1 × 14 = 14
Atomic mass of 2(O) = 2 × 16 = 32
Molecular mass of NO2 = 14 + 32 = 46 g.

(v) H2SO24
Atomic mass of 2(H) = 2 × 1= 2
Atomic mass of 1(S) = 1 × 32 = 32
Atomic mass of 4(O) = 4 × 16 = 64
Molecular mass of H2SO4 = 64 + 32 + 2 = 98 g.

Question 5.
Complete the table given below.

Element Atomic Mass Molecular Mass Atomicity
Chlorine 35.5 71
Ozone 45 3
Sulphur 32 8

Solution:

Element Atomic Mass Molecular Mass Atomicity
Chlorine 35.5 71 2
Ozone 16 48 3
Sulphur 32 256 8

Question 6.
Fill in the blanks using the given data:
The formula of Calcium oxide is CaO. The atomic mass of Ca is 40, Oxygen is 16 and Carbon is 12.

  • 1 mole of Ca (….. g) and 1 mole of the Oxygen atom (…… g) combine to form mole of CaO (….. g).
  • 1 mole of Ca (…… g) and 1 mole of C (…… g) and 3 moles of the Oxygen atom (…… g) combine to form 1 mole of CaCO3 (…… g).

Solution:

  • 1 mole of Ca (40 g) and 1 mole of the Oxygen atom (16 g) combine to form 1 mole of CaO (56 g).
  • 1 mole of Ca (40 g) and 1 mole of C (12 g) and 3 moles of the Oxygen atom (48 g) combine to form 1 mole of CaCO3 (100 g).

Question 7.
Calculate the average atomic mass of naturally occurring magnesium using the following data.
Mg – 24 = 78.99% , Mg – 25 = 10%, Mg – 26 = 11.01%
Answer:
Average atomic mass of Magnesium = atomic mass of Mg – 24 × % + atomic mass of Mg – 25 × % + atomic mass of Mg – 26 × %
= 24 × \(\frac{78.99}{100}\) + 25 × \(\frac{10}{100}\) + 26 × \(\frac{11.01}{100}\)
= 18.9576 + 2.5 + 2.8626
= = 24.3202 amu
∴ Average atomic mass of Magnesium is 24.3202 amu

Question 8.
Analyse the table and fill in the blanks.

Gas Atomic mass Molecular mass Atomicity
Ozone 16 48
Nitrogen 14 2

Solution:

Gas Atomic mass Molecular mass Atomicity
Ozone 16 48 3
Nitrogen 14 28 2

Question 9.
Analyse the table and fill in the blanks.

Substance Mass No.of moles
(a) Al 81 g
(b) Fe 0.5

Solution:

Substance Mass No.of moles
(a) Al 81 g 3
(b) Fe 27.95 g 0.5

Question 10.
When ammonia reacts with hydrogen chloride gas, it produces white fumes of ammonium chloride. The volume occupied by NH3 in glass bulb A is three times more than the volume occupied by HCl in glass bulb B at STP.
Samacheer Kalvi 10th Science Solutions Chapter 7 Atoms and Molecules 8
(i) How many moles of ammonia are present in glass bulb A?
(ii) How many grams of NH4Cl will be formed when the stopper is opened? (Atomic mass of N = 14, H = 1, Cl = 35.5)
(iii) Which gas will remain after completion of the reaction?
(iv) Write the chemical reaction involved in this process.
Solution:
(i) Capacity of NH3 bulb = 67.2 litre
22.4 litre of NH3 = 1 mole
67.2 litre of NH3 = \(\frac{1}{22.4} \times 67.2\) = 3 moles of NH3

(ii) Atomic mass of 1(N) = 1 × 14 = 14 g
Atomic mass of 4(H) = 4 × 1 = 4 g
Atomic mass of 1(Cl) = 1 × 35.5 = 35.5 g
Mass of NH4Cl = 53.5 g.

(iii) NH3 (Ammonia) gas will remain after the completion of the reaction.

(iv) Chemical equation of the reaction
NH3 (Ammonia) + HCl (Hydrochloric acid) → NH4Cl (Ammonium chloride)

Question 11.
Nitroglycerine is used as an explosive. The equation for the explosive reaction is
4C3H5((NO3))3 (l) → 12CO2 (g) + 10H2O (l) + 6N2 (g) + O2 (g)
(Atomic mass of C = 12, H = 1, N = 14, O = 16)
(i) How many moles does the equation show for:
(a) Nitroglycerine
(b) gas molecules produced?
(ii) How many moles of gas molecules are obtained from 1 mole of nitroglycerine?
(iii) What is the mass of 1 mole of nitroglycerine?
Solution:
(i) 4 moles of Nitroglycerine

(ii) 4 moles of Nitroglycerine produce 19 moles of gas molecules
1 mole of Nitroglycerine produces 19 / 4 = 4.75 moles

(iii) Mass of 1 mole of Nitroglycerine C3H5(NO3)3
Atomic mass of C = 12
Atomic mass of 3(C) = 3 × 12 = 36
Atomic mass of 5(H) = 5 × 1 = 5
Atomic mass of 3(N) = 3 × 14 = 42
Atomic mass of 9(O) = 9 × 16 = 144
Mass of 1 mole of Nitroglycerine = 227 g

Question 12.
Sodium bicarbonate breaks down on heating:
2NaHCO3 → Na2CO3 + H2O + CO2
(Atomic mass of Na = 23, C = 12, H = 1, O = 16)
(i) How many moles of sodium bicarbonate are there in this equation?
(ii) What is the mass of sodium bicarbonate used in this equation?
(iii) How many moles of carbon dioxide are there in this equation?
Solution:
\(2 \mathrm{NaHCO}_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2} \uparrow\)
(i) 2 moles of NaHCO3 (sodium bicarbonate) are there in the above equation.

(ii) Mass of sodium bicarbonate in this equation is mass of 2 moles of NaHCO3.
Atomic mass of 1(Na) = 1 × 23 = 23 g
Atomic mass of 1(H) = 1 × 1 = 1 g
Atomic mass of 1(C) = 1 × 12 = 12 g
Atomic mass of 3(O) = 3 × 16 = 48 g
Mass of 1 mole of NaHCO3 = 84 g
Mass of 2 moles of NaHCO3 = 84 × 2 = 168 g.

(iii) Number of moles of CO2 in this equation = 1 mole.

Question 13.
40 g of calcium was extracted from 56 g of calcium oxide (Atomic mass of Ca = 40, O = 16)
(i) What mass of oxygen is there in 56 g of calcium oxide?
(ii) How many moles of oxygen atoms are there in this?
(iii) How many moles of calcium atoms are there in 40 g of calcium?
(iv) What mass of calcium will be obtained from 1000 g of calcium oxide?
Solution:
(i) Mass of CaO = 56 g
Mass of Ca = 40 g
Mass of oxygen = 56 – 40 = 16 g

(ii) \(\frac{\text { No of moles of oxygen atom }}{\text { Mole }=\text { mass/atomic mass }}=\frac{16}{16}=1 \text { mole }\)
1 mole of oxygen atom.

(iii) \(\frac{\text { No of moles of calcium }}{\text { Mole }=\text { mass/atomic mass }}=\frac{40}{40}=1 \text { mole }\)
1 mole of calcium atom.

(iv) 56 g calcium oxide gives 40 g of calcium
1000 g of calcium oxide give = \(\frac{40}{56} \times 1000\)
= 714.285 g of calcium
= 714.29 g of calcium.

Question 14.
How many grams are there in the following?
(i) 1 mole of chlorine molecule, Cl2
(ii) 2 moles of sulphur molecules, S8
(iii) 4 moles of ozone molecules, O3
(iv) 2 moles of nitrogen molecules, N2
Solution:
(i) 1 mole of chlorine molecule Cl2
Atomic mass of chlorine = 35.5 g
Mass of 1 mole of chlorine = Atomic mass × Atomicity = 35.5 × 2 = 71 g.

(ii) 2 moles of sulphur molecules S8
Atomic mass of S8 = 8 × 32 = 256 g
Mass of 2 moles of S8 = Atomic mass × Number of moles = 256 × 2 = 512 g.

(iii) 4 moles of ozone molecule O3
Atomic mass of O3 = 3 × 16 = 48 g
Mass of 4 moles of ozone = 48 × 4 = 192 g.

(iv) 2 moles of Nitrogen molecule N2
Atomic mass of N2 = 2 × 14 = 28 g
Mass of 2 moles of Nitrogen = 28 × 2 = 56 g.

Question 15.
Find how many moles of atoms are there in:
(i) 2 g of nitrogen
(ii) 23 g of sodium
(iii) 40 g of calcium
(iv) 1.4 g of lithium
(v) 32 g of sulphur.
Solution:
(i) 2 g of nitrogen
Number of moles = \(\frac{\text { Mass }}{\text { atomic mass }}\)
Atomic mass of Nitrogen =14; Mass of Nitrogen = 2 g
Number of moles = \(\frac{2}{14}\) = 0.142 moles of Nitrogen.

(ii) 23 g of sodium
Atomic mass of sodium = 23
Mass of sodium = 23 g
Number of moles = \(\frac{\text { Mass }}{\text { atomic mass }}=\frac{23}{23}\) = 1 mole of sodium.

(iii) 40 g of calcium
Atomic mass of calcium = 40
Mass of calcium = 40 g
Number of moles = \(\frac{\text { Mass }}{\text { atomic mass }}=\frac{40}{40}\) = 1 mole of calcium.

(iv) 1.4 g of lithium
Atomic mass of lithium = 7
Mass of lithium = 1.4 g
Number of moles = \(\frac{\text { Mass }}{\text { atomic mass }}=\frac{1.4}{7}\) = 0.2 mole of lithium.

(v) 32g of sulphur
Atomic mass of sulphur = 32
Mass of sulphur = 32 g
Number of moles = \(\frac{\text { Mass }}{\text { atomic mass }}=\frac{32}{32}\) = 1 mole of sulphur.

SamacheerKalvi.Guru

Question 16.
Find the atomicity of chlorine, if its atomic mass is 35.5 and its molecular mass is 71.
Solution:
Atomicity = \(\frac{\text { Molecular mass }}{\text { Atomic mass }}\)
Atomicity of chlorine = \(\frac{71}{35.5}\) = 2.

Question 17.
Find the atomicity of ozone if its atomic mass is 16 and its molecular mass is 48.
Solution:
Atomicity = \(\frac{\text { Molecular mass }}{\text { Atomic mass }}\)
Atomicity of chlorine = \(\frac{48}{16}\) = 3.

Question 18.
How many atoms are present in 5 moles of oxygen?
Solution:
One mole of oxygen contains 6.023 × 1023 atoms
5 moles of oxygen contain = 5 × 6.023 × 1023
= 30.115 × 1023
= 3.0115 × 1024 atoms.

Question 19.
Calculate the number of moles in

  1. 81 g of Aluminium
  2. 4.6 g of sodium
  3. 5.1 g of ammonia
  4. 90 g of water
  5. 2 g of NaOH.

Solution:
No of moles = \(\frac{\text { Given mass }}{\text { Atomic mass }}\)

  1. No. of moles of Aluminium = \(\frac { 81 }{ 27 }\) = 3 moles of aluminium
  2. No. of moles of Sodium = \(\frac { 4.6 }{ 23 }\) = 0.2 moles of sodium
  3. No. of moles of Ammonia = \(\frac { 5.1 }{ 17 }\) = 0.3 moles of ammonia
  4. No. of moles of Water = \(\frac { 90 }{ 18 }\) = 5 moles of water
  5. No. of moles of NaOH = \(\frac { 2 }{ 40 }\) = 0.05 moles of NaOH

Question 20.
Calculate the mass of 0.5 moles of iron.
Solution:
Mass = Atomic mass × number of moles
Mass of iron = 55.9 × 0.5 = 27.95 g.

Question 21.
Find the mass of 2.5 moles of oxygen atoms.
Solution:
Mass = Atomic mass × number of moles
Mass of oxygen = 16 × 2.5 = 40 g.

Question 22.
Calculate the number of molecules in 11 g of CO2.
Solution:
Gram molecular mass of CO2 = 44 g
Number of molecules = \(\frac{\text { Avogadro number } \times \text { given mass }}{\text { Gram molecular mass }}\)
= \(\frac{6.023 \times 10^{23} \times 11}{44}\)
= 1.51 × 1023 CO2 molecules.

Question 23.
Calculate the number of molecules in 360 g of glucose.
Solution:
Number of molecules = \(\frac{\text { Avogadro number } \times \text { given mass }}{\text { Gram molar mass }}\)
Gram molar mass of glucose (C6H12O6) = (6 × 12) + (12 × 1) + (6 × 16)
= 72 + 12 + 96
= 180 g
Number of molecules = \(\frac{6.023 \times 10^{23} \times 360}{180}\)
= 6.023 × 1023 × 2
= 12.046 × 1023 molecules.
(or)
1.2046 × 1024 glucose molecules.

SamacheerKalvi.Guru

Question 24.
Calculate the mass of 18.069 × 1023 molecules of SO2?
Solution:
Mass of the substance = \(\frac{\text { Gram molecular mass } \times \text { Number of particles }}{\text { Avogadro number }}\)
Gram molecular mass of SO2 = 32 + 16(2) = 64 g
Mass of SO2 = \(\frac{64 \times 18.069 \times 10^{23}}{6.023 \times 10^{23}}\)
= 64 × 3 = 192 g.

Question 25.
Calculate the mass of glucose in 2 × 1024 molecules.
Solution:
Gram molecular mass of glucose = 180 g
Mass of glucose = \(\frac{180 \times 2 \times 10^{24}}{6.023 \times 10^{23}}\) = 597.7 g

Question 26.
Calculate the mass of 12.046 × 1023 molecules of CaO.
Solution:
Gram molecular mass of CaO = 40 + 16 = 56 g
Mass of CaO = \(\frac{56 \times 12.046 \times 10^{23}}{6.023 \times 10^{23}}\)
56 × 2 = 112 g.

Question 27.
Calculate the number of moles for a substance containing 3.0115 × 1023 molecules in it.
Solution:
Number of moles = \(\frac{\text { No. of molecules }}{\text { Avogadro number }}\)
= \(\frac{12.046 \times 10^{22}}{6.023 \times 10^{23}}\)
= 0.5 mole.

Question 28.
Calculate the number of moles in 12.046 × 1022 atoms of copper.
Solution:
No. of moles of atoms
\(\begin{array}{l}{=\frac{\text { No. of atoms }}{\text { Avogadro number }}} \\ {=\frac{12.046 \times 10^{22}}{6.023 \times 10^{23}}=2 \times 10^{-1}} \\ {=0.2 \text { mole. }}\end{array}\).

Question 29.
Calculate the number of moles in 24.092 × 1022 molecules of water.
Solution:
Number of moles =
\(\begin{array}{l}{=\frac{\text { No. of molecules }}{\text { Avogadro number }}} \\ {=\frac{24.092 \times 10^{22}}{6.023 \times 10^{23}}}\end{array}\)
= 4 × 1022 × 10-23
= 4 × 10-1
= 0.4 mole.

Question 30.
Which one of the following will have largest number of atoms?

  1. 1 g Au (s)
  2. 1 g Na (s)
  3. 1 g Li (s)
  4. 1 g of Cl2 (g)
    (Atomic masses: Au = 197, Na = 23, Li = 7, Cl = 35.5 amu)

Solution:

  1. 1 g Au = \(\frac{1}{197}\) mol = \(\frac{1}{197}\) × 6.02 × 1023 atoms
  2. 1 g Na = \(\frac{1}{23}\) mol = \(\frac{1}{23}\) × 6.02 × 1023 atoms
  3. 1 g Li = \(\frac{1}{7}\) mol = \(\frac{1}{7}\) × 6.02 × 1023 atoms
  4. 1 g Cl2 = \(\frac{1}{71}\) mol = \(\frac{1}{71}\) × 6.02 × 1023molecules = \(\frac{2}{71}\) × 6.02 × 1023 atoms.

Thus, 1 g of Li has the largest number of atoms.

Question 31.
Calculate the number of atoms in each of the following:
(i) 52 moles of He
(ii) 52 u of He
(iii) 52 g of He
Solution:
(i) 1 mol of He = 6.022 × 1023 atoms
52 mol of He = 52 × 6.022 × 1023 atoms = 3.131 × 1025 atoms.

(ii) 1 atom of He = 4 u of He
4 u of He = 1 atom of He
52 u of He = \(\frac{1}{4}\) × 52 atoms = 13 atoms.

(iii) 1 mole of He = 4 g = 6.022 × 1023 atoms
52 g of He = \(\frac{6.022 \times 10^{23}}{4} \times 52\) atoms = 7.8286 × 1024 atoms.

SamacheerKalvi.Guru

Question 32.
Calculate the number of moles in each of the following.
(i) 392 g of sulphuric acid
(ii) 44.8 litres of sulphur dioxide at N.T.P.
(iii) 6.022 × 1022 molecules of oxygen
(iv) 8 g of calcium
Solution:
(i) 392 g of sulphuric acid
Molar mass of H2SO4 = 2 × 1 + 32 + 4 × 16 = 98 g
98 g of sulphuric acid = 1 mol
392 g of sulphuric acid = 1 mol × \(\frac{392 g}{(98 g)}\) = 4 mol.

(ii) 44.8 litres of sulphur dioxide at N.T.P.
22.4 litres of sulphur dioxide at N.T.P. = 1 mol
44.8 litres of sulphur dioxide at N.T.P. = \(\frac{1 \mathrm{mol}}{(22.4 \mathrm{L})} \times(44.8 \mathrm{L})\) = 2.0 mol.

(iii) 6.022 × 1022 molecules of oxygen
6.022 × 1022 molecules of oxygen = 1 mol
6.022 × 1022 molecules of oxygen = 1 mol × \(\frac{6.022 \times 10^{22}}{6.022 \times 10^{23}}\) = 0.1 mol.

(iv) 8 g of calcium
Gram atomic mass of Ca = 40 g
40 g of calcium = 1 mol
8.0 g of calcium = 1 mol × \(\frac{(8.0 \mathrm{g})}{(40 \mathrm{g})}\) = 0.2 mol.

Question 33.
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?
Solution:
Volume of a drop of water = 0.05 mL
Mass of a drop of water = Volume × density = (0.05 mL) × (1.0 g/mL) = 0.05 g
Gram molecular mass of water (H2O) = 2 × 1 + 16 = 18 g
18 g of water = 1 mol
0.05 g of water = \(\frac{1 \mathrm{mol}}{(18 \mathrm{g})} \times(0.05 \mathrm{g})\) = 0.0028 mol
No. of molecules present
1 mole of water contain molecules = 6.022 × 1023
0.0028 mole of water contain molecules = 6.022 × 1023 × 0.0028 = 1.68 × 1021 molecules.

Question 34.
Calculate the total number of electrons present in 1.6 g of methane.
Solution:
(i) Molar mass of methane (CH4) = 12 + 4 × 1 = 16 g
16 g of methane contain molecules = 6.022 × 1023
1.6 g of methane contain molecule = \(=\frac{6.022 \times 10^{23}}{(16 \mathrm{g})} \times(1.6 \mathrm{g})\) = 6.022 × 1022

(ii) Number of electrons in 6.022 × 1022 molecules of methane
1 molecule of methane contains electrons = 6 + 4 = 10
6.022 × 1022 molecules of methane contain electrons = 6.022 × 1022 × 10 = 6.022 × 1023

Question 35.
The Vapour Density of a gaseous element is 5 times that of oxygen under similar conditions. If the molecule is triatomic, what will be its atomic mass?
Solution:
Molecular mass of oxygen = 32 u
Density of oxygen = \(\frac{32}{2}\) = 16 u
Density of gaseous element = 16 × 5 = 80 u
Molecular mass of gaseous element = 80 × 2 = 160 u
Atomicity of the element = 3
Atomic mass of the element = \(\frac{\text { Molecular mass }}{\text { Atomicity }}=\frac{160}{3}\) = 53.33 u.

Samacheer Kalvi 10th Science Solutions Chapter 9 Solutions

You can Download Samacheer Kalvi 10th Science Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 9 Solutions

Samacheer Kalvi 10th Science Solutions Textual Problems Solved

I. Problems based on Solubility.

Question 1.
1.5 g of solute is dissolved in 15 g of water to form a saturated solution at 298K. Find out the solubility of the solute at the temperature.
Solution:
Mass of the solute = 1.5 g
Mass of the solvent = 15 g
Solubility of the solute = \(\frac{\text { Mass of the solute }}{\text { Mass of the solvent }} \times 100\)
= \(\frac{1.5}{15} \times 100=10 \mathrm{g}\).

Question 2.
Find the mass of potassium chloride would be needed to form a saturated solution in 60 g of water at 303 K? Given that solubility of the KCl is 37 / 100 g at this temperature.
Solution:
Mass of potassium chloride in 100 g of water in saturated solution = 37 g
Mass of potassium chloride in 60 g of water in saturated solution = \(\frac{37}{100} \times 60\) = 22.2 g.

You can Download Samacheer Kalvi 10th Science Guide PDF help you to revise the complete Syllabus and score more marks in your examinations.

Question 3.
What is the mass of sodium chloride that would be needed to form a saturated solution in 50 g of water at 30°C. Solubility of sodium chloride is 36 g at 30°C?
Solution:
At 30°C, 36 g of sodium chloride is dissolved in 100 g of water.
Mass of sodium chloride that would be need for 100 g of water = 36 g
Mass of sodium chloride dissolved in 50 g of water = \(\frac{36 \times 50}{100}\) = 18 g.

Question 4.
The Solubility of sodium nitrate at 50°C and 30°C is 114 g and 96 g respectively. Find the amount of salt that will be thrown out when a saturated solution of sodium nitrate-containing 50 g of water is cooled from 50°C to 30°C?
Answer:
Amount of sodium nitrate dissolved in 100 g of water at 50°C is 114 g
Amount of sodium nitrate dissolving in 50 g of water at 50°C is = \(\frac{114 \times 50}{100}\) = 57 g
Similarly amount of sodium nitrate dissolving in 50 g of water at 30°C is = \(\frac{96 \times 50}{100}\) = 48 g
Amount of sodium nitrate thrown when 50 g of water is cooled from 50°C to 30°C is 57 – 48 = 9 g.

II. Problems based on Mass percentage.

Question 1.
A solution was prepared by dissolving 25 g of sugar in 100 g of water. Calculate the mass percentage of solute.
Solution:
Mass of the solute = 25 g
Mass of the solvent = 100 g
Samacheer Kalvi 10th Science Solutions Chapter 9 Solutions 9

Question 2.
16 grams of NaOH is dissolved in 100 grams of water at 25°C to form a saturated solution. Find the mass percentage of solute and solvent.
Solution:
Mass of the solute (NaOH) = 16 g
Mass of the solvent H2O = 100 g
(i) Mass percentage of the solute
= \(\frac{\text { Mass of the solute }}{\text { Mass of the solute }+\text { Mass of the solvent }} \times 100\)
= \(\frac{16 \times 100}{16+100}=\frac{1600}{116}\)
Mass percentage of the solute = 13.79 %.

(ii) Mass percentage of solvent = 100 – (Mass percentage of the solute) = 100 – 13.79 = 86.21 %.

Question 3.
Find the amount of urea which is to be dissolved in water to get 500 g of 10 % w/w aqueous solution?
Solution:
Mass percentage = \(\frac{\text { Mass of the solute }}{\text { Mass of the solution }} \times 100\)
\(10=\frac{\text { Mass of the urea }}{500} \times 100\)
Mass of urea = 50 g.

III. Problem based on Volume – Volume percentage.

Question 1.
The solution is made from 35 ml of methanol and 65 ml of water. Calculate the volume percentage.
Solution:
The volume of the ethanol = 35 ml
The volume of the water = 65 ml
Samacheer Kalvi 10th Science Solutions Chapter 9 Solutions 10

Question 2.
Calculate the volume of ethanol in 200 ml solution of 20 % v/v aqueous solution of ethanol.
Solution:
Volume of aqueous solution = 200 ml
Volume percentage = 20 %
Samacheer Kalvi 10th Science Solutions Chapter 9 Solutions 11

Activity

Activity 1.
Look at the following pictures. Label them as a dilute and concentrated solution and justify your answer.
Samacheer Kalvi 10th Science Solutions Chapter 9 Solutions 12
Solution:
In the above picture, the first teacup is more concentrated and the first beaker is more concentrated CuSO4 solution. The reason is the colour intensity is more in a concentrated solution.

Samacheer Kalvi 10th Science Solutions Textual Evaluation Solved

I. Choose the best answer.

Question 1.
A solution is a ____ mixture.
(a) homogeneous
(b) heterogeneous
(c) homogeneous and heterogeneous
(d) non – homogeneous.
Answer:
(a) homogeneous

Question 2.
The number of components in a binary solution is:
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Question 3.
Which of the following is the universal solvent?
(a) Acetone
(b) Benzene
(c) Water
(d) Alcohol.
Answer:
(c) Water

Question 4.
A solution in which no more solute can be dissolved in a definite amount of solvent at a given temperature is called:
(a) Saturated solution
(b) Unsaturated solution
(c) Supersaturated solution
(d) Dilute solution
Answer:
(a) Saturated solution

Question 5.
Identify the non – aqueous solution ______.
(a) sodium chloride in water
(b) glucose in water
(c) copper sulphate in water
(d) sulphur in carbon-di-sulphide.
Answer:
(d) sulphur in carbon-di-sulphide.

Question 6.
When pressure is increased at a constant temperature the solubility of gases in liquid:
(a) No change
(b) increases
(c) decreases
(d) no reaction
Answer:
(b) increases

Question 7.
The solubility of NaCl in 100 ml water is 36 g. If 25 g of salt is dissolved in 100 ml of water how much more salt is required for saturation ______.
(a) 12 g
(b) 11 g
(c) 16 g
(d) 20 g.
Answer:
(b) 11 g

Question 8.
A 25% alcohol solution means:
(a) 25 ml alcohol in 100 ml of water
(b) 25 ml alcohol in 25 ml of water
(c) 25 ml alcohol in 75 ml of water
(d) 75 ml alcohol in 25 ml of water
Answer:
(c) 25 ml alcohol in 75 ml of water

SamacheerKalvi.Guru

Question 9.
Deliquescence is due to ______.
(a) Strong affinity to water
(b) Less affinity to water
(c) Strong hatred of water
(d) Inertness to water.
Answer:
(a) Strong affinity to water

Question 10.
Which of the following is hygroscopic in nature?
(a) ferric chloride
(b) copper sulphate penta hydrate
(c) silica gel
(d) none of the above
Answer:
(c) silica gel

II. Fill in the blanks.

Question 1.
The component present in a lesser amount, in a solution, is called ______.
Answer:
Solute.

Question 2.
Example for liquid in solid type solution is ______.
Answer:
Sodium chloride dissolved in water.

Question 3.
Solubility is the amount of solute dissolved in _______ g of solvent.
Answer:
100.

Question 4.
Polar compounds are soluble in ______ solvents.
Answer:
Polar.

Question 5.
Volume percentage decreases with increases in temperature because of ______.
Answer:
Expansion of liquid.

III. Match the following.

Question 1.

Blue vitriol (a) CaSO4.2H2O
Gypsum (b) CaO
Deliquescence (c) CuSO4.5H2O
Hygroscopic (d) NaOH

Answer:
1 – (c), 2 – (a), 3 – (d), 4 – (b).

IV. True or False: (If false give the correct statement)

Question 1.
Solutions which contain three components are called binary solution.
Answer:
False.
Correct Statement: Solutions which contain two components are called binary solution.

Question 2.
In a solution, the component which is present in a lesser amount is called solvent.
Answer:
False.
Correct Statement:

  • In a solution, the component which is present in a larger amount is called a solvent.
  • In a solution, the component which is present in a lesser amount is called solute.

Question 3.
Sodium chloride dissolved in water forms a non-aqueous solution.
Answer:
False.
Correct Statement: Sodium chloride dissolved in water forms an aqueous solution.

Question 4.
The molecular formula of green vitriol is MgSO4.7H2O
Answer:
False.
Correct Statement: The molecular formula of green vitriol is FeSO4 .7H2O

SamacheerKalvi.Guru

Question 5.
When Silica gel is kept open, it absorbs moisture from the air, because it is hygroscopic in nature.
Answer:
True.

V. Short Answer Questions.

Question 1.
Define the term Solution.
Answer:
A solution is a homogeneous mixture of two or more substances.

Question 2.
What is mean by binary solution?
Answer:
Solutions which are made of one solute and one solvent, then it is called binary solution.

Question 3.
Give an example each

  1. gas in liquid
  2. solid in liquid
  3. solid in solid
  4. gas in gas.

Answer:

  1. Carbon – di – oxide dissolved in water (Soda water).
  2. Sodium chloride dissolved in water.
  3. Copper dissolved in gold (Alloy).
  4. A mixture of Helium – Oxygen gases.

Question 4.
What is aqueous and non-aqueous solution? Give an example.
Answer:
(i) Aqueous solution : The solution in which water acts as a solvent.
(ii) Non-aqueous solution : The solution in which any liquid other than water acts as a solvent.
Eg: Alcohol, benzene, CS2 acetone.

Question 5.
Define Volume percentage.
Answer:
Volume percentage is defined as the percentage by volume of solute (in ml) present in the given volume of the solution.
Volume percentage = \(\frac{\text { Volume of the solute }}{\text { Volume of the solution }} \times 100\).

Question 6.
The aquatic animals live more in cold region Why?
Answer:
Aquatic animals live more in cold regions because the solubility of O2 in water is more at low temperature and therefore the amount of dissolved O2 is more in the water of cold regions.

Question 7.
Define Hydrated salt.
Answer:
The number of water molecules found in the crystalline substance or salts is called water of crystallization. Such salts are called hydrated salts.

Question 8.
A hot saturated solution of copper sulphate forms crystals as it cools. Why?
Answer:
A hot saturated solution of CuSO4 forms crystal as it cools. Because on cooling the water molecules move closer together and there is less space for the solution to hold on to as much of the dissolved solid and so it forms crystals.

Question 9.
Classify the following substances into deliquescent, hygroscopic. Conc. Sulphuric acid, Copper sulphate penta hydrate, Silica gel, Calcium chloride, and Gypsum salt.
Answer:

  1. Deliquescent substances: Calcium chloride
  2. Hygroscopic substances: Conc Sulphuric acid, Copper sulphate penta hydrate, Silica gel and Gypsum salt.

VI. Long Answer Questions.

Question 1.
Write notes on
(i) saturated solution
(ii) unsaturated solution
Answer:
(i) Saturated solution : A solution in which no more solute can be dissolved in a definite amount of the solvent at a given temperature is called saturated solution.
Eg: 36 g of sodium chloride in 100 g of water at 25°C forms saturated solution.

(ii) Unsaturated solution : Unsaturated solution is one that contains less solute than that of the saturated solution at a given temperature.
Eg: 10 g or 20 g or 30 g of Sodium chloride in 100 g of water at 25°C forms an unsaturated solution.

Question 2.
Write notes on various factors affecting solubility.
Answer:
Factors affecting solubility: There are three main factors which govern the solubility of the solute. They are
(i) Nature of the solute and solvent
(ii) Temperature
(iii) Pressure

(i) Nature of the solute and solvent: The nature of the solute and solvent plays an important role insolubility. Although water dissolves an enormous variety of substances, both ionic and covalent, it does not dissolve everything. The phrase that scientists often use when predicting solubility is “like dissolves like.” This expression means that dissolving occurs when similarities exist between the solvent and the solute. For example, Common salt is a polar compound and dissolves readily in polar solvent like water.

Non – polar compounds are soluble in non-polar solvents. For example, Fat dissolved in ether. But non-polar compounds, do not dissolve in polar solvents; polar compounds do not dissolve in non-polar solvents.

(ii) Effect of Temperature
The solubility of Solids in Liquid: Generally, the solubility of a solid solute in a liquid solvent increases with increase in temperature. For example, a greater amount of sugar will dissolve in warm water than in cold water.
In the endothermic process, solubility increases with increase in temperature. In the exothermic process, solubility decreases with increase in temperature.

The solubility of Gases in liquid: Solubility of gases in liquids decreases with increase in temperature. Generally, water contains dissolved oxygen. When water is boiled, the solubility of oxygen in water decreases, so oxygen escapes in the form of bubbles. Aquatic animals live more in cold regions because more amount of dissolved oxygen is present in the water of cold regions. This shows that the solubility of oxygen in water is more at low temperatures.

(iii) Effect of Pressure: Effect of pressure is observed only in the case of solubility of a gas in a liquid. When the pressure is increased, the solubility of a gas in liquid increases.
The common examples for solubility of gases in liquids are carbonated beverages, i.e. soft drinks, household cleaners containing an aqueous solution of ammonia, formalin aqueous solution of formaldehyde, etc.
Samacheer Kalvi 10th Science Solutions Chapter 9 Solutions 1

Question 3.
(a) What happens when MgSO4.7H2O is heated? Write the appropriate equation.
(b) Define solubility.
Answer:
(a) MgSO4.7H2O has a water of crystallization is 7. When magnesium sulphate heptahydrate crystals are gently heated, it loses seven water molecules and becomes anhydrous magnesium sulphate.
Samacheer Kalvi 10th Science Solutions Chapter 9 Solutions 2

(b) Solubility: It is defined as the number of grams of a solute that can be dissolved in 100 g of a solvent to form its saturated solution at a given temperature and pressure.

Question 4.
In what way, hygroscopic substances differ from deliquescent substances.
Answer:
Difference between hygroscopic and deliquescent substances is in the extent to which each material can absorb moisture. This is because both of these terms are very much related to each other and they refer to the property of observing and the retention of moisture from the air. However, they differ in the extent of absorption of moisture where hygroscopic materials absorb moisture but not to the extent the original substance dissolves in it, which is the case with deliquescence. Therefore deliquescence can be regarded as an extreme condition of hygroscopic activity.

Difference between hygroscopic substances and deliquescence

Hygroscopic Substances Deliquescence Substances
When exposed to the atmosphere at ordinary temperature, they absorb moisture and do not dissolve. When exposed to the atmospheric air at ordinary temperature, they absorb moisture and dissolve.
Hygroscopic substances do not change their physical state on exposure to air. Deliquescent substances change its physical state on exposure to air.
Hygroscopic substances may be amorphous solids or liquids. Deliquescent substances are crystalline solids.

Question 5.
A solution is prepared by dissolving 45 g of sugar in 180 g of water. Calculate the mass percentage of solute.
Answer:
Mass of sugar (solute) = 45 g
Mass of water (solvent) = 180 g.
Samacheer Kalvi 10th Science Solutions Chapter 9 Solutions 3

Question 6.
3.5 litres of ethanol is present in 15 litres of an aqueous solution of ethanol. Calculate volume per cent of the ethanol solution.
Answer:
Volume of ethanol (solute) = 3.5 litre
Volume of aqueous solution = 15 litre
Volume of the solution = \(\frac{\text { Volume of the solute }}{\text { Volume of the solution }} \times 100\)
= \(\frac{3.5}{15} \times 100\)
Volume percentage = 23.33 %.

VII. HOT Questions.

Question 1.
Vinu dissolves 50 g of sugar in 250 ml of hot water, Sarath dissolves 50 g of same sugar in 250 ml of cold water. Who will get faster dissolution of sugar? and Why?
Answer:
Vinu will get faster dissolution of sugar. Because Vinu dissolves 50 g of sugar in 250 ml of hot water, whereas Sarath dissolves 50 g of sugar in 250 ml of cold water. Solubility of a solid in liquid increases with increase in temperature.

Question 2.
‘A’ is a blue coloured crystalline salt. On heating, it loses a blue colour and to give ‘B’. When water is added, ‘B’ gives back to ‘A’. Identify A and B, write the equation.
Answer:

  • Blue coloured crystalline salt is copper sulphate pentahydrate (A)
  • On heating Copper sulphate pentahydrate it loses blue colour and to give anhydrous copper sulphate (B).
  • When water is added to the anhydrous copper sulphate (B) gives back to copper sulphate pentahydrate (A).
  • Samacheer Kalvi 10th Science Solutions Chapter 9 Solutions 4

Question 3.
Will the cool drinks give more fizz at top of the hills or at the foot? Explain.
Answer:
Carbonated cool drinks give more fizz at the foot hill because solubility of gases in liquid decrease with increase in temperature. At higher altitudes, the temperature is low and the dissolved CO2 will not escape as fizz, whereas at the foot hill it does.

Samacheer Kalvi 10th Science Solutions Additional Questions Solved

I. Choose the best answer.

Question 1.
Find out the homogeneous mixture.
(a) Salt + Water
(b) Sand + Water
(c) Clay + Water
(d) Gold + Water.
Answer:
(a) Salt + Water

Question 2.
The aqueous solution is:
(a) S in CS2
(b) I2 in CCl4
(c) Salt in H2O
(d) None of the above
Answer:
(c) Salt in H2O

Question 3.
In a solution, the component which is present in a larger amount is called ______.
(a) solvent
(b) dissolution
(c) solute
(d) mole.
Answer:
(а) solvent

Question 4.
The solubility of ammonia gas at 25°C is:
(a) 36 g
(b) 48 g
(c) 80 g
(d) 184 g
Answer:
(b) 48 g

Question 5.
Example for liquid in liquid binary solution is ______.
(a) Copper dissolved in gold
(b) Water vapour in the air
(c) Ethyl alcohol dissolved in water
(d) NaCl dissolved in water.
Answer:
(c) Ethyl alcohol dissolved in water

Question 6.
Which one of the following is an aqueous solvent?
(a) Benzene
(b) Acetone
(c) Alcohol
(d) Water.
Answer:
(d) Water.

Question 7.
Mass percentage of a solution is independent of:
(a) temperature
(b) amount of solute
(c) amount of solvent
(d) chemical nature of the solute
Answer:
(a) temperature

Question 8.
Solubility is equal to ______.
Samacheer Kalvi 10th Science Solutions Chapter 9 Solutions 5
Answer:
(c) \(\frac{\text { Mass of the solute }}{\text { Mass of the solvent }} \times 100\)

Question 9.
In which case solubility increases with increase in temperature?
(a) Endothermic process
(b) Exothermic process
(c) Both (a) and (b)
(b) None of these.
Answer:
(a) Endothermic process

Question 10.
The factor/factors which affect the solubility of a solute:
(a) Nature of the solute and solvent
(b) Temperature
(c) Pressure
(d) All of these
Answer:
(d) All of these

Question 11.
The effect of pressure on the solubility of a gas in a liquid is given by ______.
(a) Hess’s law
(b) Ohm’s law
(c) Henry’s law
(d) Gases law.
Answer:
(c) Henry’s law

Question 12.
5% sugar solution means ______.
(a) 5 g of sugar in 95 g of water
(b) 50 g of sugar in 50 g of water
(c) 20 g of sugar in 80 g of water
(d) 95 g of sugar in 5 g of water.
Answer:
(a) 5 g of sugar in 95 g of water

Question 13.
The water of crystallisation present in white vitriol is:
(a) 2
(b) 7
(c) 5
(d) 24
Answer:
(b) 7

Question 14.
Mass percentage is independent of ______.
(a) density
(b) volume
(c) pressure
(d) temperature.
Answer:
(d) temperature.

Question 15.
Volume percentage is expressed as ______.
(a) v/v
(b) w/w
(c) v/w
(d) w/v.
Answer:
(a) v/v

Question 16.
When the temperature increases, volume percentage ______.
(a) Increases
(b) Decreases
(c) no change
(d) increases then decrease.
Answer:
(b) Decreases

Question 17.
The mass of NaCl needed to form saturated solution in 50 g of water at 30°C. When the solubility of NaCl is 36 g at 30°C is:
(a) 9 g
(b) 18 g
(c) 57 g
(d) 19 g
Answer:
(b) 18 g

Question 18.
Zinc sulphate heptahydrate is also called as ______.
(a) Blue Vitriol
(b) Epsom salt
(c) Green Vitriol
(d) White Vitriol.
Answer:
(d) white Vitriol.

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Question 19.
Water of crystallization of blue vitriol is ______.
(a) 1
(b) 5
(c) 4
(d) 2.
Answer:
(b) 5

Question 20.
On heating copper sulphate pentahydrate crystals ______.
(a) blue colour changes into green
(b) green colour changes into blue
(c) blue colour changes into colourless
(d) no colour changes.
Answer:
(c) blue colour changes into colourless

Question 21.
Water of crystallization of Epsom salt is ______.
(a) 1
(b) 5
(c) 4
(d) 1.
Answer:
(d) 7.

Question 22.
Hygroscopic substances are used as ______.
(a) drying agents
(b) hydrating agents
(c) freezing agents
(d) reducing agents.
Answer:
(a) drying agents

Question 23.
If two liquids are mutually soluble, they are called _____ liquids.
(a) miscible
(b) immiscible
(c) insoluble
(d) viscous.
Answer:
(a) miscible

Question 24.
Aquatic species are more comfortable in cold water because of ______.
(a) as the temperature decreases, the solubility of dissolved oxygen increases
(b) as the temperature increases, the solubility of dissolved oxygen increases
(c) as the temperature increases, the solubility of dissolved oxygen decreases
(d) None of these.
Answer:
(a) as the temperature decreases, the solubility of dissolved oxygen increases

Question 25.
The process of food assimilation by man is in the form of ______.
(a) solution
(b) solid
(c) solute
(d) solvent.
Answer:
(a) solution

Question 26.
Common salt in water is an example of ______.
(a) Colloidal solution
(b) Binary solution
(c) Suspension
(d) all the above.
Answer:
(b) Binary solution

Question 27.
36 g of NaCl in 100 g of water is a ____ solution.
(a) Non – aqueous
(b) Unsaturated
(c) Supersaturated
(d) Saturated.
Answer:
(d) Saturated.

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Question 28.
Nitrogen in soil is an example of _____ solution in nature.
(a) saturated
(b) unsaturated
(c) supersaturated
(d) aqueous.
Answer:
(a) saturated

Question 29.
The solubility of CuSO4 in H2O is ______ at 20°C.
(a) 36 g
(b) 20.7 g
(c) 10 g
(d) 95 g.
Answer:
(b) 20.7 g

Question 30.
The solubility of NaNO3 in 100 g water at 25°C is ______.
(a) 36 g
(b) 95 g
(c) 184 g
(d) 92 g.
Answer:
(d) 92 g.

Question 31.
Which of the following affect solubility?
(a) Temperature
(b) Nature of solute and solvent
(c) Pressure
(d) all the above.
Answer:
(d) all the above.

Question 32.
In _____ process, solubility increases with increase in temperature.
(a) exothermic
(b) endothermic
(c) isothermic
(d) adiabatic.
Answer:
(b) endothermic

Question 33.
In ______ process, solubility decreases with increase in temperature.
(a) exothermic
(b) adiabatic
(c) endothermic
(d) isothermic.
Answer:
(a) exothermic

Question 34.
Increase in ______ increases the solubility of gases.
(a) temperature
(b) pressure
(c) no. of moles
(d) concentration.
Answer:
(b) pressure

Question 35.
The solubility of NaBr in H2O is ______.
(a) 36 g
(b) 95 g
(c) 184 g
(d) 92 g.
Answer:
(b) 95 g

Question 36.
The gas – filled in soft drinks in ______.
(a) O2
(b) N2
(c) CO2
(d) H2
Answer:
(c) CO2

Question 37.
The solubility of KNO3 in water is ______ process.
(a) Exothermic
(b) endothermic
(c) Isothermic
(d) Adiabatic.
Answer:
(b) endothermic

Question 38.
The solubility of CaO in water is ______ process.
(a) Exothermic
(b) endothermic
(c) Adiabatic
(d) Isothermic.
Answer:
(a) Exothermic

Question 39.
The solubility of NaI in water is ______.
(a) 184 g
(b) 92 g
(c) 95 g
(d) 36 g.
Answer:
(a) 184 g

II. Fill in the blanks.

Question 1.
A solution is a _____ mixture of two or more substances.
Answer:
Homogeneous.

Question 2.
Common salt dissolved in water is an example for _____ solution.
Answer:
Binary.

Question 3.
In a solution, the component present in a lesser amount by weight is called ____ and in a large amount by weight is called ______.
Answer:
Solute, solvent.

Question 4.
The solution in which water acts as a solvent is called ______ and the solution in which Benzene acts as a solvent is called ______.
Answer:

  1. The aqueous solution,
  2. Non – aqueous solution.

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Question 5.
20 g of NaCl in 100 g of water is _____ solution.
Answer:
Unsaturated.

Question 6.
36 g of NaCl in 100 g of water at room temperature forms a ____ solution.
Answer:
Saturated.

Question 7.
Nitrogen in soil is an example for _____ in nature.
Answer:
Saturated solution.

Question 8.
A solution which has more of solute than the saturated solution at a given temperature is called ______ solution.
Answer:
Supersaturated.

Question 9.
In _____ process, solubility increases with ______ in temperature.
Answer:
Endothermic, increase.

Question 10.
In an ______ process, solubility decreases with ______ in temperature.
Answer:
Exothermic, increase.

Question 11.
The solubility of KNO3 increases with the _____ in temperature.
Answer:
Increase.

Question 12.
The solubility of CaO decreases with the ______ in temperature.
Answer:
Increase.

Question 13.
The solubility of oxygen is _____ in cold water.
Answer:
More.

Question 14.
A polar compound is insoluble in _____ solvent.
Answer:
Non – polar.

Question 15.
An increase in ______ increases the solubility of a gas in a liquid.
Answer:
Pressure.

Question 16.
______ gas is filled in soft drinks using the effect of pressure.
Answer:
CO2

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Question 17.
Hygroscopic substances are used as ______.
Answer:
Drying agents.

Question 18.
Anhydrous calcium chloride is an example of ______.
Answer:
hygroscopic substance.

Question 19.
Hygroscopic substances absorb moisture without changing their ______.
Answer:
Physical state.

Question 20.
Deliquescent substances lose their ______.
Answer:
Crystalline shape.

Question 21.
Ferric chloride is an example of ______.
Answer:
Deliquescent substances.

Question 22.
On heating, hydrated crystalline salts lose their ______.
Answer:
Water of crystallization.

Question 23.
The number of water molecules in blue vitriol is ______.
Answer:
Five.

Question 24.
The number of water molecules in magnesium sulphate heptahydrate is ______.
Answer:
Seven.

Question 25.
Mass percentage is _______ to temperature.
Answer:
Independent.

Question 26.
To qualify the solute in a solution, we can use the term ______.
Answer:
Concentration.

Question 27.
Based on the amount of solute, the solution is classified into ________ types.
Answer:
Three.

Question 28.
Zinc dissolved in copper is an example of ______ solution.
Answer:
Solid in solid.

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Question 29.
______ is an example of gas in the liquid.
Answer:
Carbon dioxide dissolved in water.

Question 30.
______ and _______ are in the form of solution to decide the physiological activity of human beings.
Answer:
Blood, Lymph.

III. Match the following.

Question 1.

i. Sugar in water (a) Exothermic process
ii. Sulphur in carbon sulphide (b) Aqueous solution
iii. The solubility of CaO (c) Endothermic process
iv. Solubility of KNO3 (d) Non-aqueous solution

Answer:
i – b, ii – d, iii – a, iv – c.

Question 2.

i. Saturated solution (a) Sulphur in acetone
ii. Unsaturated solution (b) 56 g NaCl in water
iii. Supersaturated solution (c) 10 g of NaCl in water
iv. Non-aqueous solution (d) 36 g of NaCl in water

Answer:
i – d, ii – c, iii – b, iv – a.

Question 3.

i. Solubility of KNO3 (a) decreases with increase in temperature
ii. The solubility of CaO (b) increases with increase in pressure
iii. The solubility of oxygen (c) increases with increase in temperature
iv. Solubility of CO2 (d) more in cold water

Answer:
i – c, ii – a, iii – d, iv – b.

Question 4.

i. Mercury with sodium (a) Liquid in gas
ii. NaCl in water (b) Liquid in solid
iii. CO2 in water (c) Solid in liquid
iv. Water vapour in air (d) Gas in liquid

Answer:
i – b, ii – c, iii – d, iv – a.

Question 5.

Solute Solubility
i. CaCO3 (a) 80
ii. NH3 (b) 91
iii. NaOH (c) 0.0013
iv. C6H12O6 (d) 48

Answer:
i – c, ii – d, iii – a, iv – b.

Question 6.

i. Epsom salt (a) ZnSO4.7H2O
ii. Green Vitriol (b) MgSO4.7H2O
iii. White Vitriol (c) SiO2
iv. Silica gel (d) FeSO4.7H2O

Answer:
i – b, ii – d, iii – a, iv – c.

IV. State whether true or false. If false, give the correct statement.

Question 1.
A solution is a heterogeneous mixture of two or more substances.
Answer:
False.
Correct statement: A solution is a homogeneous mixture of two or more substances.

Question 2.
A solution contains four components, it is called as a binary solution.
Answer:
False.
Correct statement: A solution contains two components, it is called as a binary solution.

Question 3.
The solution in which water acts as a solvent, it is called a non-aqueous solution.
Answer:
False.
Correct statement: The solution in which water acts as a solvent, it is called an aqueous solution.

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Question 4.
The solution of sulphur in CS2 is a suitable example of a non-aqueous solution.
Answer:
True.

Question 5.
Nitrogen in soil is an example of a supersaturated solution.
Answer:
False.
Correct statement: Nitrogen in the soil is an example of a saturated solution.

Question 6.
The solubility of CuSO4 in H2O is 36 g at 25°C.
Answer:
False.
Correct statement: Solubility of CuSO4 in H2O is 20.7 g at 20°C.

Question 7.
100 ml of water can dissolve 36 g of NaCl at 25°C to attain saturation.
Answer:
True.

Question 8.
In an endothermic process, solubility decreases with increase in temperature.
Answer:
False.
Correct statement: In an endothermic process, solubility increases with increase in temperature.

Question 9.
In an exothermic process, solubility decreases with increase in temperature.
Answer:
True.

Question 10.
The solubility of CaO increases with the increase in temperature.
Answer:
False.
Correct statement: The solubility of CaO decreases with the increase in temperature.

Question 11.
The solubility of oxygen is more in cold water.
Answer:
True.

Question 12.
SO2 gas is filled in soft drinks using the effect of pressure.
Answer:
False.
Correct statement: CO2 gas is filled in soft drinks using the effect of pressure.

Question 13.
Ether act as a universal solvent.
False.
Correct statement: Water acts universal solvent.

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Question 14.
Covalent compounds are soluble in water.
Answer:
False.
Correct statement: Covalent compounds are insoluble in water.

Question 15.
36 g of sodium chloride in 100 g of water at 25°C forms a saturated solution.
Answer:
True.

Question 16.
40 g of sodium chloride in 100 g of water at 25°C forms an unsaturated solution.
Answer:
False.
Correct statement: 40 g of sodium chloride in 100 g of water at 25°C forms a supersaturated solution.

Question 17.
Aquatic animals live more in hot regions.
Answer:
False.
Correct statement: Aquatic animals live more in cold regions.

Question 18.
Mass percentage is dependent of temperature.
Answer:
False.
Correct statement: Mass percentage is independent of temperature.

Question 19.
Hygroscopic substances do not change their physical state on exposure to air.
Answer:
True.

Question 20.
Deliquescent substances do not change their physical state on exposure to air.
Answer:
False.
Correct statement: Deliquescent substances change its physical state on exposure to air.

V. Assertion and Reason

Question 1.
Assertion (A): Salt solution-common salt dissolved in water is an example of a binary solution.
Reason (R): A solution with two components is called a binary solution.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 2.
Assertion (A): Sugar in water is a true solution.
Reason (R): True solution is a homogeneous mixture that contains small solute particles that are dissolved throughout the solvent.
(a) Both (A) and (R) are wrong
(b) Both (A) and (R) are correct
(c) (A) is wrong but (R) is correct
(d) (A) is correct but (R) is wrong.
Answer:
(b) Both (A) and (R) are correct

Question 3.
Assertion (A): Solution of sulphur in CS2 is an example of a non-aqueous solution.
Reason (R): The solution in which any liquid other than water acts as a solvent is called the non – aqueous solution.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

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Question 4.
Assertion (A): Solubility of KNO3 increases with increase in temperature.
Reason (R): Solubility of KNO3 is an endothermic process.
(a) Both (A) and (R) are wrong
(b) Both (A) and (R) are correct
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct

Question 5.
Assertion (A): Aquatic species are more comfortable in cold water.
Reason (R): Solubility of oxygen is more in cold water.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 6.
Assertion (A): CO2 gas is filled in soft drinks using the effect of pressure.
Reason (R): A decrease in pressure increases the solubility of a gas in a liquid.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(c) (A) is correct but (R) is wrong

Question 7.
Assertion (A): Aqueous copper sulphate is a binary solution.
Reason (R): Copper sulphate solution contains two components i.e., one solute-copper sulphate and one solvent-water.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 8.
Assertion (A): Water act as a universal solvent.
Reason (R): Most of the substances are soluble in water.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 9.
Assertion (A): When Blue vitriols are gently heated it turns colourless.
Reason (R): It loses five water molecules and becomes anhydrous compounds.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 10.
Assertion (A): Anhydrous calcium chloride is an example of a hygroscopic substance.
Reason (R): Anhydrous calcium chloride changes its physical state on exposure to air.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is wrong but (R) is correct
(d) (A) is correct but (R) is wrong
Answer:
(d) (A) is correct but (R) is wrong.

VI. Short Answer Questions.

Question 1.
State Henry’s Law.
Answer:
The solubility of a gas in liquid is directly proportional to the pressure of the gas over the solution at a definite temperature.

Question 2.
Define solute and solvent.
Answer:
In a solution, the component present in a lesser amount by weight is called solute and the component present in a large amount by weight is called solvent.

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Question 3.
What is a
(i) saturated and
(ii) supersaturated solution?
Answer:
(i) Saturated : A solution in which no more solute can be dissolved in a definite amount of the solvent at a given temperature.
(ii) Super saturated solution : A solution which contains more solute than the saturated solution at a given temperature.

Question 4.
Why water is known as a universal solvent?
Answer:
Most of the solutes (substances) are soluble in water and so it is called universal solvent.

Question 5.
Common salt dissolves in water easily. Give reason.
Answer:
Common salt is an electrolyte. It is easily dissociated into its ions like Na+ and Cl in polar solvent water. So NaCl (common salt) an inorganic compound readily dissolves in water.

Question 6.
What is meant by the ternary solution?
Answer:
Solutions which contain three components are called ternary solutions.
E.g. If salt and sugar are added to water, both dissolves in water forming a solution. Here two solutes are dissolved in one solvent. Therefore it is a ternary solution.

Question 7.
Name the type of solution formed in the following cases;

  1. 20 g of NaCl in 100 g of water at 25°C
  2. 36 g of NaCl in 100 g of water at 25°C
  3. Nitrogen in soil
  4. Sulphur in CS2.

Answer:

  1. Unsaturated
  2. Saturated
  3. Saturated
  4. Non-aqueous solution

Question 8.
What are concentrated and dilute solutions?
Answer:
Two solutions having the same solute and solvent, the one which contains a higher amount of solute per the given amount of solvent is said to be concentrated solution and the other is said to be dilute solution.

Question 9.
Define Henry’s law.
Answer:
To quantify the solute in a solution, we can use the term concentration. The concentration of a solution may be defined as the amount of solute present in a given amount of solution or solvent.
(or)
The solubility of a gas in a liquid is directly proportional to the pressure of the gas over the solution at a definite temperature.

Question 10.
Define Mass percentage.
Answer:
Mass percentage of a solution is defined as the percentage by mass of the solute present in the solution.
Mass percentage,
\(=\frac{\text { Mass of the solute }}{\text { Mass of the solution }} \times 100\).

Question 11.
What is water crystallization?
Answer:
The number of water molecules found in the crystalline substance is called water of crystallization.

Question 12.
What are hygroscopic substances?
Answer:
Certain substances, when exposed to the atmospheric air at ordinary temperature, absorb moisture without changing their physical state. Such substances are called hygroscopic substances and this property is called hygroscopy.

Question 13.
List out the examples for hygroscopic substances.
Answer:

  • Conc.Sulphuric acid (H2SO4).
  • Phosphorus Pentoxide (P2O5).
  • Quick lime (CaO).
  • Silica gel (SiO2).
  • Anhydrous calcium chloride (CaCl2).

Question 14.
What are the Deliquescent substances?
Answer:
Certain substances which are so hygroscopic, when exposed to the atmospheric air at ordinary temperatures, absorb enough water and get completely dissolved. Such substances are called deliquescent substances and this property is called deliquescence.

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Question 15.
When is deliquescence is maximum?
Answer:

  • The temperature is low
  • The atmosphere is humid.

Question 16.
Give examples for deliquescent substances.
Answer:

  • Calcium chloride (CaCl2)
  • Caustic soda (NaOH)
  • Caustic potash (KOH)
  • Ferric chloride (FeCl3).

VII. Long Answer Questions.

Question 1.
From the table given below, furnish your points of inference.

Substance Solubility at 25°C
NaCl 36 g
NaBr 95 g
NaI 184 g

Answer:
Inferences:

  • At 25°C, 36g NaCl is dissolved in 100 g water to give a saturated solution.
  • At 25°C, 95 g NaBr is dissolved in 100 g water to get a saturated solution.
  • At 25°C, 184g Nal is dissolved in 100 g water to get a saturated solution.
  • The solubility of a solute at a given solvent at a particular temperature is defined as the number of grams of solute necessary to saturate 100 g of the solvent at that temperature.
  • In the above tabular column, we infer that the solubility of NaI is the highest and the solubility of NaCl is the lowest.
  • All the NaCl, NaBr, Nal solution in water are called aqueous solutions.
  • Depending upon the amount of the solute, the solutions are classified as a saturated and unsaturated solution.
  • The above solutions are saturated solutions.

Question 2.
Distinguish between the saturated and unsaturated solution at a temperature of 25°C using the data given below (Note: Solubility of NaCl is 36 g).

  1. 16 g NaCl in 100 g water
  2. 36 g NaCl in 100 g water.

Answer:
The solubility of NaCl is 36g.

  1. 16 g NaCl in 100 g water is an unsaturated solution.
  2. 36 g NaCl in 100 g water is a saturated solution.
Saturated Solution Unsaturated Solution
1. A solution in which no more solute can be dissolved in a definite amount of the solvent at a given temperature is called a saturated solution. 1. A solution in which the solute is lesser in the amount in the solvent is called unsaturated solution.
2. In this solution, no more solute can be dissolved. If more of solute is added, it will not dissolve to give a supersaturated solution. 2. More of solute can be dissolved to get a saturated solution.

Question 3.
Write a note on the type of solution based on the amount of solute present in a solution.
Answer:
Based on the amount of solute, in the given amount of solvent, solutions are classified into the following types.
(i) Saturated solution : A solution in which no more solute can be dissolved in a definite amount of the solvent at a given temperature is called saturated solution. Eg: 36 g of sodium chloride in 100 g of water at 25°C forms saturated solution.Further addition of sodium chloride, leave it undissolved.

(ii) Unsaturated solution : Unsaturated solution is one that contains less solute than that of the saturated solution at a given temperature. Eg: 10 g or 20 g or 30 g of Sodium chloride in 100 g of water at 25°C forms an unsaturated solution.

(iii) Super saturated solution : Supersaturated solution is one that contains more solute than the saturated solution at a given temperature. Eg: 40 g of sodium chloride in 100 g of water at 25°C forms super saturated solution. This state can be achieved by altering any other conditions liken temperature, pressure. Super saturated solutions are unstable, and the solute is reappearing as crystals when the solution is disturbed.

Question 4.
Find the concentration of a solution in terms of weight per cent if 20 g of common salt is dissolved in 50 g of water.
Answer:
Samacheer Kalvi 10th Science Solutions Chapter 9 Solutions 6

VIII. HOT Questions.

Question 1.
Observe the diagram.

  1. Which is a concentrated solution and why?
  2. Which is dilute solution and why?
    Samacheer Kalvi 10th Science Solutions Chapter 9 Solutions 7

Answer:

  1. Flask (b) is a concentrated solution. Because (b) flask contains a large number of solute particles than (a) flask.
  2. Flask (a) is a dilute solution. Because (a) flask contains a lesser number of solute particles than (b) flask.

Question 2.
Where we use the phrase – “Like dissolves Like” and explain the meaning of the phrase?
Answer:
The phrase “like dissolves like” is often used for predicting solubility. This expression means that dissolving occurs when similarities exist between the solvent and the solute. For example, Common salt is a polar compound and dissolves readily in polar solvent like water.

Question 3.
Why bubbling occurs when water is boiled?
Answer:
The solubility of a gas in liquid decrease with increase in temperature. Generally, water contains dissolved oxygen. When water is boiled, the solubility of oxygen in water decreases, so oxygen escapes in the form of bubbles.

Question 4.
What happens when blue vitriol is heated? Explain.
Answer:
When blue vitriol (Copper sulphate pentahydrate) crystals are gently heated, it loses it five water molecules and becomes colourless anhydrous copper sulphate.
Samacheer Kalvi 10th Science Solutions Chapter 9 Solutions 8

Samacheer Kalvi 10th Science Solutions Additional Problems Solved

Question 1.
Take 10 g of common salt and dissolve it in 40 g of water. Find the concentration of a solution in terms of weight per cent.
Solution:
Weight percent = \(\frac{\text { Weight of the solute }}{\text { Weight of solute }+\text { Weight of solvent }} \times 100\)
\(=\frac{10}{10+40} \times 100=20 \%\).

Question 2.
2 g of potassium sulphate was dissolved in 12.5 ml of water. On cooling, the first crystals appeared at 60°C. What is the solubility of potassium sulphate in water at 60°C? Solution. 12.5 ml of water weighs 12.5 g.
Solution:
In 12.5 g of water, the amount of potassium sulphate dissolved is 2 g.
In 1 g of water, the amount of potassium sulphate dissolved is 2 / 12.5 g.
Hence in 100 g of water, the amount of potassium sulphate dissolved is (2 × 100) / 12.5 = 16 g.
The solubility of potassium sulphate in the water at 60°C is 16 g.

Question 3.
50 g of a saturated solution of NaCl at 30°C is evaporated to dryness and 13.2 g of dry NaCl was obtained. Find the solubility of NaCl at 30°C in water.
Solution:
Mass of water in solution = 50 – 13.2 = 36.8 g
Solubility of NaCl
\(\begin{aligned} &=\frac{\text { Mass of } \mathrm{NaCl}}{\text { Mass of water }} \times 100 \\=& \frac{13.2}{36.8} \times 100=36 \mathrm{g} \end{aligned}\)
Solubility of NaCl = 36 g (approx).

Question 4.
An empty evaporating dish weighs 20.0 g. After adding a saturated solution of NaNO3, the dish weighs 66.0 g. When evaporated to dryness, the dish with crystals weighs 42.5 g. Find the solubility of NaNO3 at 20°C.
Solution:
Weight of saturated solution of NaNO3 = (66.0 – 20.0) g = 46.0 g
Weight of crystals of NaNO3 = (41.5 – 20.0) g = 21.5 g
Weight of water in saturated solution = (46.0 – 21.5) g = 24.5 g
Solubility of NaNO3 = \(\frac{\text { Weight of NaNO }_{3} \text { crystals }}{\text { Weight of water }} \times 100\)
\(=\frac{21.5}{24.5} \times 100=87.7 \mathrm{g}\)
Solubility of NaNO3 at 20°C is = 87.7 g in 100 g H2O.

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Question 5.
30 g sugar is dissolved in 170g of water. Calculate the concentration of solution in terms of weight percent.
Solution:
Weight percent = \(\frac{\text { Weight of the solute }}{\text { Weight of solute }+\text { Weight of solvent }} \times 100\)
\(=\frac{30}{30+170} \times 10=15 \%\).

Question 6.
50 g common salt is dissolved in 150 g of water; Find oat the concentration of solution in terms of weight percent.
Solution:
Weight percent = \(\frac{\text { Weight of the solute }}{\text { Weight of solute }+\text { Weight of solvent }} \times 100\)
\(=\frac{50}{50+150} \times 100\)
= 25 %.

Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.1

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.1

Question 1.
Find all positive integers which when divided by 3 leaves remainder 2.
Answer:
The positive integers when divided by 3 leaves remainder 2.
By Euclid’s division lemma a = bq + r, 0 ≤ r < b.
Here a = 3q + 2, where 0 ≤ q < 3, a leaves remainder 2 when divided by 3.
∴ 2, 5, 8, 11 ……………..

Question 2.
A man has 532 flower pots. He wants to arrange them in rows such that each row contains 21 flower pots. Find the number of completed rows and how many flower pots are left over?
Answer:
Here a = 532, b = 21
Using Euclid’s division algorithm
a = bq + r
532 = 21 × 25 + 7
Number of completed rows = 21
Number of flower pots left over = 7

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Question 3.
Prove that the product of two consecutive positive integers is divisible by 2.
Solution:
Let n – 1 and n be two consecutive positive integers. Then their product is (n – 1)n.
(n – 1)(n) = n2 – n.
We know that any positive integer is of the form 2q or 2q + 1 for some integer q. So, following cases arise.
Case I. When n = 2q.
In this case, we have
n2 – n = (2q)2 – 2q = 4q2 – 2q = 2q(2q – 1)
⇒ n2 – n = 2r, where r = q(2q – 1)
⇒ n2 – n is divisible by 2.

Case II. When n = 2q + 1
In this case, we have
n2 – n = (2q + 1)2 – (2q + 1)
= (2q + 1)(2q + 1 – 1) = 2q(2q + 1)
⇒ n2 – n = 2r, where r = q (2q + 1).
⇒ n2 – n is divisible by 2.
Hence, n2 – n is divisible by 2 for every positive integer n.
Hence it is Proved

Question 4.
When the positive integers a, b and c are divided by 13, the respective remainders are 9,7 and 10. Show that a + b + c is divisible by 13.
Answer:
Let the positive integer be a, b, and c
We know that by Euclid’s division lemma
a = bq + r
a = 13q + 9 ….(1)
b = 13q + 7 ….(2)
c = 13q + 10 ….(3)
Add (1) (2) and (3)
a + b + c = 13q + 9 + 13q + 7 + 13q + 10
= 39q + 26
a + b + c = 13 (3q + 2)
This expansion will be divisible by 13
∴ a + b + c is divisible by 13

Question 5.
Prove that square of any integer leaves the remainder either 0 or 1 when divided by 4.
Solution:
Let x be any integer.
The square of x is x2.
Let x be an even integer.
x = 2q + 0
then x2 = 4q2 + 0
When x be an odd integer
When x = 2k + 1 for some interger k.
x2 = (2k + 1 )2
= 4k2 + 4k + 1
= 4k (k + 1) + 1
= 4q + 1
where q = k(k + 1) is some integer.
Hence it is proved.

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Question 6.
Use Euclid’s Division Algorithm to find the Highest Common Factor (H.C.F) of
(i) 340 and 412
Answer:
To find the HCF of 340 and 412 using Euclid’s division algorithm. We get
412 = 340 × 1 + 72
The remainder 72 ≠ 0
Again applying Euclid’s division algorithm to the division of 340
340 = 72 × 4 + 52
The remainder 52 ≠ 0
Again applying Euclid’s division algorithm to the division 72 and remainder 52 we get
72 = 52 × 1 + 20
The remainder 20 ≠ 0
Again applying Euclid’s division algorithm
52 = 20 × 2 + 12
The remainder 12 ≠ 0
Again applying Euclid’s division algorithm
20 = 12 × 1 + 8
The remainder 8 ≠ 0
Again applying Euclid’s division algorithm
12 = 8 × 1 + 4
The remainder 4 ≠ 0
Again applying Euclid’s division algorithm
8 = 4 × 2 + 0
The remainder is zero
∴ HCF of 340 and 412 is 4

(ii) 867 and 255
Answer:
To find the HCF of 867 and 255 using
Euclid’s division algorithm. We get
867 = 255 × 3 + 102
The remainder 102 ≠ 0
Using Euclid’s division algorithm
255 = 102 × 2 + 51
The remainder 51 ≠ 0
Again using Euclid’s division algorithm
102 = 51 × 2 + 0
The remainder is zero
∴ HCF = 51
∴ HCF of 867 and 255 is 51

(iii) 10224 and 9648
Answer:
Find the HCF of 10224 and 9648 using Euclid’s division algorithm. We get
10224 = 9648 × 1 + 576
The remainder 576 ≠ 0
Again using Euclid’s division algorithm
9648 = 576 × 16 + 432
The remainder 432 ≠ 0
Using Euclid’s division algorithm
576 = 432 × 1 + 144
The remainder 144 ≠ 0
Again using Euclid’s division algorithm
432 = 144 × 3 + 0
The remainder is 0
∴ HCF = 144
The HCF of 10224 and 9648 is 144

(iv) 84,90 and 120
Answer:
Find the HCF of 84, 90 and 120 using Euclid’s division algorithm
90 = 84 × 1 + 6
The remainder 6 ≠ 0
Using Euclid’s division algorithm
4 = 14 × 6 + 0
The remainder is 0
∴ HCF = 6
The HCF of 84 and 90 is 6
Find the HCF of 6 and 120
120 = 6 × 20 + 0
The remainder is 0
∴ HCF of 120 and 6 is 6
∴ HCF of 84, 90 and 120 is 6

Question 7.
Find the largest number which divides 1230 and 1926 leaving remainder 12 in each case.
Solution:
The required number is the H.C.F. of the numbers.
1230 – 12 = 1218,
1926 – 12 = 1914
First we find the H.C.F. of 1218 & 1914 by Euclid’s division algorithm.
1914 = 1218 × 1 + 696
The remainder 696 ≠ 0.
Again using Euclid’s algorithm
1218 = 696 × 1 + 522
The remainder 522 ≠ 0.
Again using Euclid’s algorithm.
696 = 522 × 1 + 174
The remainder 174 ≠ 0.
Again by Euclid’s algorithm
522 = 174 × 3 + 0
The remainder is zero.
∴ The H.C.F. of 1218 and 1914 is 174.
∴ The required number is 174.

Question 8.
If d is the Highest Common Factor of 32 and 60, find x and y satisfying d = 32x + 60y.
Answer:
Find the HCF of 32 and 60
60 = 32 × 1 + 28 ….(1)
The remainder 28 ≠ 0
By applying Euclid’s division lemma
32 = 28 × 1 + 4 ….(2)
The remainder 4 ≠ 0
Again by applying Euclid’s division lemma
28 = 4 × 7 + 0….(3)
The remainder is 0
HCF of 32 and 60 is 4
From (2) we get
32 = 28 × 1 + 4
4 = 32 – 28
= 32 – (60 – 32)
4 = 32 – 60 + 32
4 = 32 × 2 -60
4 = 32 x 2 + (-1) 60
When compare with d = 32x + 60 y
x = 2 and y = -1
The value of x = 2 and y = -1

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Question 9.
A positive integer when divided by 88 gives the remainder 61. What will be the remainder when the same number is divided by 11?
Solution:
Let a (+ve) integer be x.
x = 88 × y + 61
61 = 11 × 5 + 6 (∵ 88 is multiple of 11)
∴ 6 is the remainder. (When the number is divided by 88 giving the remainder 61 and when divided by 11 giving the remainder 6).

Question 10.
Prove that two consecutive positive integers are always coprime.
Answer:

  1. Let the consecutive positive integers be x and x + 1.
  2. The two number are co – prime both the numbers are divided by 1.
  3. If the two terms are x and x + 1 one is odd and the other one is even.
  4. HCF of two consecutive number is always 1.
  5. Two consecutive positive integer are always coprime.

Fundamental Theorem of Arithmetic
Every composite number can be written uniquely as the product of power of prime is called fundamental theorem of Arithmetic.

Samacheer Kalvi 10th Science Solutions Chapter 15 Nervous System

You can Download Samacheer Kalvi 10th Science Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 15 Nervous System

Samacheer Kalvi 10th Science Nervous System Textual Evaluation Solved

I. Choose the Correct Answer.

Question 1.
Bipolar neurons are found in _______
(a) retina of eye
(b) cerebral cortex
(c) embryo
(d) respiratory epithelium
Answer:
(a) retina of eye

Question 2.
Site for processing of vision, hearing,memory, speech, intelligence and thought is:
(a) kidney
(b) ear
(c) brain
(d) lungs
Answer:
(c) brain

You can Download Samacheer Kalvi 10th Science Guide PDF help you to revise the complete Syllabus and score more marks in your examinations.

Question 3.
In reflex action, the reflex arc is formed by _______
(a) brain, spinal cord, muscle
(b) receptor, muscle, spinal cord
(c) muscle, receptor, brain
(d) receptor, spinal cord, muscle
Answer:
(b) receptor, muscle, spinal cord

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Question 4.
Dendrites transmit impulse ……cell body and axon transmit impulse …….. cell body.
(a) away from, away from
(b) towards, away from
(c) towards, towards
(d) away from, towards
Answer:
(b) towards, away from

Question 5.
The outer most of the three cranial meninges is ______
(a) arachnoid membrane
(b) piamater
(c) duramater
(d) myelin sheath
Answer:
(c) duramater

Question 6.
There are ______ pairs of cranial nerves and ______ pairs of spinal nerves.
(a) 12, 31
(b) 31, 12
(c) 12, 13
(d) 12, 21
Answer:
(a) 12, 31

Question 7.
The neurons which carries impulse from the central nervous system to the muscle fibre:
(a) afferent neurons
(b) association neuron
(c) efferent neuron
(d) unipolar neuron
Answer:
(c) efferent neuron

Question 8.
Which nervous band connects the two cerebral hemispheres of brain?
(a) thalamus
(b) hypothalamus
(c) corpus callosum
(d) pons
Answer:
(c) corpus callosum

Question 9.
Node of Ranvier is found in ______
(a) muscles
(b) axons
(c) dendrites
(d) cyton
Answer:
(b) axons

Question 10.
Vomiting centre is located in:
(a) medulla oblongata
(b) stomach
(c) cerebrum
(d) hypothalamus
Answer:
(a) medulla oblongata

Question 11.
Nerve cells do not possess _______
(a) neurilemma
(b) sarcolemma
(c) axon
(d) dendrites
Answer:
(b) sarcolemma

Question 12.
A person who met with an accident lost control of body temperature, water balance, and hunger. Which of the following part of brain is supposed to be damaged?
(a) Medulla oblongata
(b) cerebrum
(c) pons
(d) hypothalamus
Answer:
(d) hypothalamus

II. Fill in the blanks.

Question 1.
______ is the longest cell in our body.
Answer:
Axon

Question 2.
Impulses travels rapidly in ______ neurons.
Answer:
Myelin sheath of

Question 3.
A change in the environment that causes an animal to react is called ______
Answer:
reactions or responses

Question 4.
_____ carries the impulse towards the cell body.
Answer:
Dendrites

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Question 5.
The two antagonistic components of the autonomic nervous system are _____ and ______
Answer:
Sympathetic, Parasympathetic

Question 6.
A neuron contains all cell organelles except _______
Answer:
Golgi apparatus in axon

Question 7.
_____ maintains the constant pressure inside the cranium.
Answer:
Cerebrospinal fluid

Question 8.
______ and ______ increases the surface area of cerebrum.
Answer:
Gyri and Sulci

Question 9.
The part of the human brain which acts as a relay centre is _______
Answer:
Thalamus

III. State whether True or False, if false write the correct statement.

Question 1.
Dendrons are the longest fibres that conduct impulses away from the cell body.
Answer:
False
Correct Statement: Axons are the longest fibres that conduct impulses away from the cell body.

Question 2.
The sympathetic nervous system is a part of the central nervous system.
Answer:
False
Correct Statement: Sympathetic nervous system is a part of the autonomic nervous system.

Question 3.
Hypothalamus is the thermoregulatory centre of the human body.
Answer:
True

Question 4.
The cerebrum controls the voluntary actions of our body.
Answer:
True

Question 5.
In the central nervous system, myelinated fibres form the white matter.
Answer:
False
Correct Statement: In the central nervous system, two types of matter such as white matter or grey matter, is formed, with respect to the presence or absence of myelin sheath.

Question 6.
All the nerves in the body are covered and protected by meninges.
Answer:
False
Correct Statement: The brain is covered by three connective tissue membrane or meninges.

Question 7.
Cerebrospinal fluid provides nutrition to brain.
Answer:
True

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Question 8.
Reflex arc allows the rapid response of the body to a stimulus.
Answer:
True

Question 9.
Pons helps in regulating respiration.
Answer:
True

IV. Match the following

Question 1.

1. Nissil’s granules (a) Forebrain
2. Hypothalamus (b) Peripheral Nervous system
3. Cerebellum (c) Cyton
4. Schwann cell (d) Hindbrain

Answer:
1. (c) Cyton
2. (a) Forebrain
3. (d) Hindbrain
4. (b) Peripheral Nervous system

V. Understand the assertion statement. Justify the reason given and choose the correct choice.

(a). Assertion is correct and Reason is wrong
(b). Reason is correct and the assertion is wrong
(c). Both assertion and reason are correct
(d). Both assertion and reason are wrong

Question 1.
Assertion: Cerebrospinal fluid is present throughout the central nervous system.
Reason: Cerebrospinal fluid has no such functions.
Answer:
(a) Assertion is correct and Reason is wrong

Question 2.
Assertion: Corpus callosum is present in space between the dura mater and pia mater. Reason: It serves to maintain constant intracranial pressure.
Answer:
(d) Both assertion and reason are wrong

VI. Short Answer Questions.

Question 1.
Define the stimulus.
Answer:
The changes in the environmental condition, that are detected by receptors present in the body are called stimulus.

Question 2.
Name the parts of the hind brain.
Answer:
Hind brain consists of cerebellum, pons and medulla oblongata.

Question 3.
What are the structures involved in the protection of the brain?
Answer:
The brain is covered by three connective tissue membrane or meninges.

  • Dura mater, which is the outermost thick fibrous membrane.
  • Arachnoid membrane, which is the middle thin vascular membrane providing a web-like cushion.
  • Pia mater, which is the innermost, thin delicate membrane richly supplied with blood. Meningeal membranes protect the brain from mechanical injury.

Question 4.
Give an example for conditioned reflexes.
Answer:
Common examples of conditioned reflexes are playing a musical instrument, tying shoelaces or the neck-tie without being attentive, watering of the mouth after seeing or smelling favourite food.

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Question 5.
Which acts as a link between the nervous system and the endocrine system?
Answer:
The hypothalamus controls the secretion of hormones from the Anterior Pituitary gland and is an important link between the nervous system and the endocrine system.

Question 6.
Define reflex arc.
Answer:
The path along which the reflexive impulse travel is called reflex arc.
(OR)
The path way taken by nerve impulse to accomplish reflex action is called reflex arc.

VII. Differentiate between

Question 1.
Voluntary and involuntary actions.
Answer:

Voluntary Action Involuntary Action
1. The actions which are under the control of our will, eg. Eating, walking. 1. The actions, which are not under our control, eg. Breathing, Heartbeat
2. Controlled by the brain. 2. Controlled by the spinal cord.
3. Voluntary Action results in muscular action. 3. Involuntary actions result in a muscular action or secretions of some glands.

Question 2.
Medullated and non-medullated nerve fibre.
Answer:

Medullated nerve fibre Non-medullated nerve fibre
1. Nerve fibre is covered by a protective sheath, called the Myelin sheath, which is covered by Neurilemma. 1. Nerve fibre is covered by a single sheath, Neurilemma.
2. Nodes of Ranvier are present. 2. Nodes of Ranvier are absent.
3. They appear white. 3. They appear grey.
4. They carry nerve impulses, much faster than non-medullated nerve fibre. 4. They carry nerve impulses, much slower than medullated nerve fibre.
5. They are present in the white matter of brain, spinal cord and in the cranial and spinal nerves. 5. They are present in the grey matter of the brain and spinal cord and in the autonomic nerves.

VIII. Long Answer Questions.

Question 1.
With a neat labelled diagram explain the structure of a neuron
Answer:
Structure of Neuron: A neuron consists of three basic parts namely Cyton, Dendrites and Axon.
Samacheer Kalvi 10th Science Solutions Chapter 15 Nervous System 1
(a) Cyton: Cyton is called cell body or perikaryon. It has a central nucleus with abundant cytoplasm called neuroplasm. The cytoplasm has a large granular body called Nissl’s granules and the other cell organelles like mitochondria, ribosomes, lysosomes, and endoplasmic reticulum. Neurons do not have the ability to divide. Several neurofibrils are present in the cytoplasm that help in the transmission of nerve impulses to and from the cell body.

(b) Dendrites: These are the numerous branched cytoplasmic processes, that project from the surface of the cell body. They conduct nerve impulses, towards the cyton. The branched projections increase the surface area for receiving the signals from other nerve cells.

(c) Axon: The axon is a single, elongated, slender projections. The end of axon terminates as fine branches, which terminate into knob like swellings called synaptic knob.

The plasma membrane of axon is called axolemma, while the cytoplasm is called axoplasm. It carries impulses away from the cyton. The axons may be covered by a protein sheath called myelin sheath, which is further covered by a layer of Schwann cells called neurilemma.

Myelin sheath breaks at intervals, by depressions called Nodes of Ranvier. The region between the nodes is called an internode. Myelin sheath acts as an insulator and ensures the rapid transmission of nerve impulses.
A junction between synaptic knob of the axon of one neuron and dendron of next neuron is called Synaptic Junction. Information from one neuron can pass to another neuron through these junctions, with the release of chemicals known as neurotransmitters, from the synaptic knob.

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Question 2.
Illustrate the structure and functions of the brain.
Answer:
A human brain is formed of three main parts forebrain, midbrain and hindbrain.
1. Forebrain: The forebrain is formed of Cerebrum and Diencephalon. The diencephalon consists of dorsal thalamus and ventral hypothalamus.
(a) Cerebrum: It is the largest portion and nearly forms two-third of the brain. The cerebrum is longitudinally divided into two halves, as of right and left cerebral hemispheres by a deep – cleft called median cleft. The two cerebral hemispheres are interconnected by thick band of nerve fibres called corpus callosum. The outer portion of each cerebral hemisphere is formed of grey matter and is called cerebral cortex.

The inner or deeper part is formed of white matter and is called cerebral medulla. The cortex is extremely folded forming elevations called gyri, with depressions between them termed as sulci, that increase the surface area. Each cerebral hemisphere is divisible into a frontal lobe, a parietal lobe, a temporal lobe and an occipital lobe. These lobes are also known as cerebral lobes.

(b) Thalamus: Thalamus present in the cerebral medulla is a major conducting centre for sensory and motor signalling. It acts as a relay centre.

(c) Hypothalamus: It lies at the base of the thalamus. It controls the secretions of hormones from the anterior pituitary gland.

2. Midbrain: It is located between thalamus and Hindbrain. The dorsal portion of the midbrain consists of four rounded bodies called corporaquadrigemina, that control visual and auditory (hearing) reflexes.

3. Hindbrain: It is formed of three parts Cerebellum, Pons and Medulla Oblongata.

  • Cerebellum: It is the second-largest part of the brain formed of two large-sized hemispheres and middle vermis.
  • Pons: It is a bridge of nerve fibre that connects the lobes of the cerebellum. It relays signals between the cerebellum, spinal cord, midbrain and cerebrum.
  • Medulla Oblongata: It is the posterior-most part of the brain, that connects the spinal cord and various parts of the brain.

Functions of Brain:

Structure Functions
1. Cerebral cortex Sensory perception, Intelligence, consciousness, control of voluntary functions, language, thinking, memory, decision making, creativity, reasoning and will power.
2. Thalamus Acts as Relay Station.
3. Hypothalamus Temperature control, anger, thirst, hunger, urination, the important link between the nervous system and endocrine glands, sleep, sweating, sexual desire, fear, water balance, blood pressure.
4. Midbrain Visual and Auditory reflexes.
5. Cerebellum Maintenance of posture and balance, and co-ordinate voluntary muscle activity.
6. Pons Respiration and Role in the sleep-wake cycle.
7. Medulla Oblongata Cardiovascular, respiratory and digestive control centres, vasomotor centres to control heartbeat, contraction of blood vessels. It also regulates vomiting and salivation.

Question 3.
What will you do if someone pricks your hand with a needle? Elucidate the pathway of response with a neat labelled diagram.
Answer:
Samacheer Kalvi 10th Science Solutions Chapter 15 Nervous System 2
The pathway taken by nerve impulse to accomplish reflex action is called the reflex arc.

  • When the hand is pricked by a needle, the stimulus is the prick (touch) pain, which is sensed by a receptor called touch or pain receptors in the skin in our hand. This stimulus in tum triggers an impulse in a sensory neuron.
  • The sensory neuron transmits or conveys the message to the spinal cord.
  • The spinal cord interprets the stimulus and the impulse is passed on to the relay neuron, which in tum transmits it to a motor neuron.
  • Motor neurons carry command from the spinal cord to our arm.
  • The muscle in our arm contracts and we withdraw our hand immediately from the needle prick. Muscle is the effector organ which has responded to the prick (pain).

Question 4.
Describe the structure of the spinal cord.
Answer:
The spinal cord is a cylindrical structure lying in the neural canal of the vertebral column. It is also covered by meninges. It extends from the lower end of medulla oblongata to the first lumbar vertebra. The posterior-most region of spinal cord tapers into a thin fibrous thread-like structure called Filum terminate.

Internally, the spinal cord contains a cerebrospinal fluid-filled cavity, known as the central canal. The grey matter of the spinal cord is ‘H’ shaped. The upper end of the letter, ‘H’ forms posterior horns and lower end forms anterior horns. A bundle of fibres pass into the posterior horn forming the dorsal or afferent root. Fibres pass outward, from the anterior horn forming the ventral or efferent root. These two roots joins to form spinal nerves. The white matter is external and has a bundle of nerve tracts. Spinal cord conducts sensory and motor impulses to and from the brain. It controls the reflex actions of the body.

Question 5.
How nerve impulses are transferred from one neuron to next neuron?
Answer:
All the information from the environment is detected by the receptors, located in the sense organs such as the eyes, nose, skin and etc. Information from the receptors is transmitted as electrical impulse and is received by the dendritic tips of the neuron. This impulse travels from the dendrite to the cell body and then along the axon to its terminal end. On reaching the axonal end, it causes the nerve endings to release a chemical (neurotransmitter) which diffuses across a synapse and starts a similar electrical impulse in the dendrites of the next neuron, then to their cell body to be carried along the axon.

The electrical signal reaches the brain or spinal cord. The response from the brain (or spinal cord) is similarly passed on to the effector organs such as the muscle or gland cell, that undergoes the desired response.

Question 6.
Classify neurons based on its structure.
Answer:
The neurons are classified, based on their structures:
Samacheer Kalvi 10th Science Solutions Chapter 15 Nervous System 3
(a) Unipolar neurons: Only one nerve process arises from the cyton, which acts as both axon and dendron. It is found in early embryos, but not in the adult.

(b) Bipolar neurons: The cyton gives rise to two nerve processes, of which one acts as an axon, while another acts as a dendron. Bipolar neurons are found in the retina of the eye and olfactory epithelium of nasal chambers.

(c) Multipolar neurons: The cyton gives rise to many dendrons and an axon. Multipolar neurons found in the cerebral cortex of the brain.

IX. Higher Order Thinking Skills (HOTS) Questions

Question 1.
‘A’ is a cylindrical structure that begins from the lower end of medulla and extend downwards. It is enclosed in bony cage ‘B’ and covered by membranes ‘C’ As many as ‘D’ pairs of nerves arise from the structure ‘A’.

  1. What is A?
  2. Name (a) bony cage ‘B’ and (b) membranes ‘C’
  3. How much is D?

Answer:

  1. A is spinal cord.
  2. (a) Bony cage is Vertebral column
    (c) is Meninges
  3. D is 31 pairs of spinal nerve.

Question 2.
Our body contains a large number of cells ‘L’ which are the longest cells in the body. L has a long and short branch called as ‘M’ and ‘N’ respectively. There is a gap ‘O’ between two ‘L’ cells, through which nerve impulse transfer by the release of chemical substance ‘P’.

  1. Name the cells L
  2. What are M and N?
  3. What is gap O?
  4. Name the chemical substance P?

Answer:

  1. Neuron
  2. Axon and Dendrites
  3. Synaptic Junction (knob)
  4. Neurotransmitter (Acetylcholine)

Samacheer Kalvi 10th Science Nervous System Additional Questions Solved

I. Fill in the blanks.

Question 1.
The condition needed for the coordination between the various cells and organ for the diverse activities is called _____
Answer:
Homeostasis

Question 2.
A number of nerve fibres bundled up together to form _____
Answer:
Nerves

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Question 3.
The other name for cyton is cell body or _____
Answer:
Perikaryon

Question 4.
______ is an important neurotransmitter released by neurons.
Answer:
Acetylcholine

Question 5.
_______ membranes protect the brain from mechanical injury.
Answer:
Meningeal

Question 6.
The four rounded bodies of the midbrain are called ______
Answer:
Corporaquadrigemina

Question 7.
The Posterior most region of spinal cord tapers into fibrous thread-like structures called _____
Answer:
Filum terminale

II. Choose the incorrect statement and Write it as a correct statement.

Question 1.
Responses refer to the changes in the environmental conditions, which are detected by receptors.
Answer:
Incorrect Statement
Correct Statement: Stimulus refer to the changes in the environmental conditions, which are detected by receptors. .

Question 2.
The brain is the controlling centre of all the body activities.
Answer:
Correct Statement

Question 3.
Axon carries impulses towards the cyton.
Answer:
Incorrect Statement
Correct Statement: Axon carries impulses away from the cyton.

Question 4.
CNS consists of all nerves, which connect the brain and spinal cord to all parts of the body.
Answer:
Incorrect Statement
Correct Statement: CNS consists of all nerves, which connect the brain and spinal cord to all parts of the body.

Question 5.
A receptor is a cell or group of cells that receive the stimuli.
Answer:
Correct Statement

III. Match the following.

Question 1.

1. Neuroglia (a) Conduct nerve impulses
2. Electrical impulse (b) Connects the lobe of the cerebellum
3. Pons (c) Cerebrospinal fluid
4. Brain (d) Do not conduct nerve impulses
5. Dendrites (e) Information from the receptor

Answer:
1. (d) Do not conduct nerve impulses
2. (e) Information from the receptor
3. (b) Connects the lobe of the cerebellum
4. (c) Cerebrospinal fluid
5. (a) Conduct nerve impulses

IV. Answer the following in a word or with a Sentence.

Question 1.
What are Spinal Reflexes?
Answer:
Most of the reflex actions are monitored and controlled by the spinal cord. So it is called Spinal Reflexes.

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Question 2.
Name the two enlargement found in spinal cord.
Answer:
The spinal cord has two enlargements.
Cervical plexus – found in neck region
Lumbar plexus – in Lumbar region.

Question 3.
What is the function of Meningeai membranes in the brain?
Answer:
Meningeal membranes protect the brain from mechanical injury.

Question 4.
Write a function of glial cells.
Answer:
Glial cells or Neuroglia are non-exciting, supporting cells of nervous system.

Question 5.
Name the granules in the cytoplasm of Peyton.
Answer:
Nissl’s granules.

Question 6.
Which is the controlling centre of all body activities?
Answer:
Brain

Question 7.
In which fluid, is the brain suspended?
Answer:
The brain is suspended in the cerebrospinal fluid.

Question 8.
Name the four rounded bodies in the dorsal portion of midbrain.
Answer:
Corpora quadrigemina

Question 9.
Which is the longest cell of the human body?
Answer:
The longest cell of the human body is axon.

Question 10.
Name the tapered spinal cord into a thin thread-like structure present in the posterior-most region.
Answer:
Filum terminate.

V. Draw a neat labelled diagram of the following.

Question 1.
(a) Brain
(b) Structure of Spinal Cord
(c) Nerve impulse transmission
Answer:
(a) Brain
Samacheer Kalvi 10th Science Solutions Chapter 15 Nervous System 4
(b) Structure of Spinal Cord
Samacheer Kalvi 10th Science Solutions Chapter 15 Nervous System 5
(c) Nerve impulse transmission
Samacheer Kalvi 10th Science Solutions Chapter 15 Nervous System 6

VI. Answer the following briefly.

Question 1.
What is the synaptic transmission?
Answer:
The flow of nerve impulses from the axonal end of one neuron to the dendrite of another neuron through a synapse is called synaptic transmission.

Question 2.
What is homeostasis?
Answer:
The coordination between the various cells and organs is essential for their diverse activities to maintain physiological balance called Homeostasis.

Question 3.
What are Neurotransmitters? Give an example.
Answer:
Neurotransmitters are chemicals, which allow the transmission of a nerve impulse from the axon terminal of one neuron to the dendron of another neuron or to an effector organ. Acetylcholine is the neurotransmitter released by neurons.

Question 4.
Name the three connective tissue membrane of Meninges.
Answer:

  1. Duramater : Outermost thick fibrous membrane.
  2. Arachnoid membrane : Middle thin vascular membrane provide web like cushion.
  3. Piamater : Innermost thin oblicate membrane richly supplied with blood.

Question 5.
What are the types of nerve fibres?
Answer:
The two types of nerve fibres, based on the presence or absence of myelin sheath are

  • Myelinated nerve fibre: The axon is covered with a myelin sheath.
  • Non-myelinated nerve fibre: The axon is not covered by a myelin sheath.

Myelinated and non-myelinated nerve fibres from the white matter and grey matter of the brain.

Question 6.
Why is the Autonomic Nervous System called Visceral nervous system?
Answer:
The Autonomic Nervous System is also called Visceral Nervous System because it regulates the function of internal visceral organs of our body.

Question 7.
What are the cerebral lobes?
Answer:
Each cerebral hemisphere is divisible into a frontal lobe, a parietal lobe, a temporal lobe and an occipital lobe. These lobes are called cerebral lobes.

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Question 8.
How are neurons categorized on the basis of functions?
Answer:
On the basis of functions the neurons are categorised as:

  • Sensory or afferent neurons, which carry impulses from the sense organ to the central nervous system.
  • Motor or efferent neurons, which carry impulses from the central nervous system to effector organs such as the muscle fibre and the gland.
  • Association neurons, which conduct impulses between sensory and motor neurons.

Question 9.
What is cerebrospinal fluid? What is its functions?
Answer:
The brain is suspended in a special fluid environment called cerebrospinal fluid (CSF).
Functions:

  • It acts as a shock-absorbing fluid and protects the brain from damage when it is subjected to a sudden jerk.
  • It supplies nutrients to the brain.
  • It collects and removes wastes from the brain.
  • It is responsible for maintaining a constant pressure inside the cranium.

Question 10.
Explain the Peripheral Nervous System.
Answer:
Peripheral Nervous System is formed, by the nerves arising from the brain and the spinal cord. The nerves arising from the brain are called cranial nerves. In man, there are 12 pairs of cranial nerves. Some of the cranial nerves are sensory.
Nerves arising from the spinal cord are called spinal nerves. In man, there are 31 pairs of spinal nerves. Each spinal nerve has a dorsal sensory root and the ventral motor root.

VI. Higher Order Thinking Skills (HOTS) Questions

Question 1.
Raju got an injury on the head during motorbike accident on the road. Later he faced a problem in maintaining balance of the body while walking or sitting. Which part of the brain do you think is affected?
Answer:
Cerebellum in the brain is affected as it maintains the body balance.

Question 2.
What is Electroencephalogram (EEG)?
Answer:
Electroencephalogram (EEG) is an instrument, which records the electrical impulses of the brain. An EEG can detect abnormalities in the brain waves and help in diagnoses of epilepsy, brain tumours and head injuries, etc.

Question 3.
Name a few brain diseases.
Answer:

  • Alzheimer’s disease (develop, as we age)
  • Parkinson’s disease
  • Amyotrophic Lateral Sclerosis (ALS)

Question 4.
What are brain injuries? Give examples.
Answer:
The damage of brain tissue, neurons and nerves. This damage affects our brain’s ability to communicate with the rest of our body. Examples of Brain injuries include blood clots, hematomas, swelling inside the skull, strokes and etc.

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Question 5.
What are the symptoms of brain injury?
Answer:
Vomiting, Nausea, Speech difficulty, bleeding from the ear, Numbness and etc.

Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Ex 1.4

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Ex 1.4

Question 1.
Determine whether the graph given below represent functions. Give reason for your answers concerning each graph.
Samacheer Kalvi 10th Maths Chapter 1 Relations and Functions Ex 1.4 1
Solution:
Samacheer Kalvi 10th Maths Chapter 1 Relations and Functions Ex 1.4 2
(i) It is not a function. The graph meets the vertical line at more than one points.
(ii) It is a function as the curve meets the vertical line at only one point.
(iii) It is not a function as it meets the vertical line at more than one points.
(iv) It is a function as it meets the vertical line at only one point.

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Question 2.
Let f :A → B be a function defined by f(x) = \(\frac{x}{2}\) – 1, Where A = {2, 4, 6, 10, 12},
B = {0, 1, 2, 4, 5, 9}. Represent f by
(i) set of ordered pairs;
(ii) a table;
(iii) an arrow diagram;
(iv) a graph
Solution:
f: A → B
A = {2, 4, 6, 10, 12}, B = {0, 1, 2, 4, 5, 9}
Samacheer Kalvi 10th Maths Chapter 1 Relations and Functions Ex 1.4 3
(i) Set of ordered pairs
= {(2, 0), (4, 1), (6, 2), (10, 4), (12, 5)}
(ii) a table
Samacheer Kalvi 10th Maths Chapter 1 Relations and Functions Ex 1.4 4
(iii) an arrow diagram;
Samacheer Kalvi 10th Maths Chapter 1 Relations and Functions Ex 1.4 5

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Question 3.
Represent the function f = {(1, 2),(2, 2),(3, 2), (4,3), (5,4)} through
(i) an arrow diagram
(ii) a table form
(iii) a graph
Solution:
f = {(1, 2), (2, 2), (3, 2), (4, 3), (5, 4)}
(i) An arrow diagram.
Samacheer Kalvi 10th Maths Chapter 1 Relations and Functions Ex 1.4 6

Question 4.
Show that the function f : N → N defined by f{x) = 2x – 1 is one – one but not onto.
Solution:
f: N → N
f(x) = 2x – 1
N = {1, 2, 3, 4, 5,…}
f(1) = 2(1) – 1 = 1
f(2) = 2(2) – 1 = 3
f(3) = 2(3) – 1 = 5
f(4) = 2(4) – 1 = 7
f(5) = 2(5) – 1 = 9
Samacheer Kalvi 10th Maths Chapter 1 Relations and Functions Ex 1.4 7
In the figure, for different elements in x, there are different images in f(x).
Hence f : N → N is a one-one function.
A function f: N → N is said to be onto function if the range of f is equal to the co-domain of f
Range = {1, 3, 5, 7, 9,…}
Co-domain = {1, 2, 3,..}
But here the range is not equal to co-domain. Therefore it is one-one but not onto function.

Question 5.
Show that the function f: N → N defined by f (m) = m2 + m + 3 is one – one function.
Solution:
f: N → N
f(m) = m2 + m + 3
N = {1, 2, 3, 4, 5…..}m ∈ N
f{m) = m2 + m + 3
f(1) = 12 + 1 + 3 = 5
f(2) = 22 + 2 + 3 = 9
f(3) = 32 + 3 + 3 = 15
f(4) = 42 + 4 + 3 = 23
Samacheer Kalvi 10th Maths Chapter 1 Relations and Functions Ex 1.4 8
In the figure, for different elements in the (X) domain, there are different images in f(x). Hence f: N → N is a one to one but not onto function as the range of f is not equal to co-domain.
Hence it is proved.

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Question 6.
Let A = {1,2, 3, 4} and B = N .
Let f: A → B be defined by f(x) = x3 then,
(i) find the range of f
(ii) identify the type of function
Answer:
A = {1,2, 3,4}
B = {1,2, 3, 4, 5,….}
f(x) = x3
f(1) = 13 = 1
f(2) = 23 = 8
f(3) = 33 = 27
f(4) = 43 = 64
(i) Range = {1,8, 27, 64}
(ii) one -one and into function.

Question 7.
In each of the following cases state whether the function is bijective or not. Justify your answer.
(i) f: R → R defined by f(x) = 2x + 1
(ii) f: R → R defined by f(x) = 3 – 4x2
Solution:
(i) f : R → R
f(x) = 2x + 1
f(1) = 2(1) + 1 = 3
f(2) = 2(2) + 1 = 5
f(-1) = 2(-1) + 1 = -1
f(0) = 2(0) + 1 = 1
It is a bijective function. Distinct elements of A have distinct images in B and every element in B has a pre-image in A.
(ii) f: R → R; f(x) = 3 – 4x2
f(1) = 3 – 4(12) = 3 – 4 = -1
f(2) = 3 – 4(22) = 3 – 16 = -13
f(-1) = 3 – 4(-1)2 = 3 – 4 = -1
It is not bijective function since it is not one-one

Question 8.
Let A = {-1, 1} and B = {0, 2}. If the function f: A → B defined by f(x) = ax + b is an onto function? Find a and b.
Solution:
A= {-1, 1},B = {0, 2}
f: A → B, f(x) = ax + b
f(-1) = a(-1) + b = -a + b
f(1) = a(1) + b = a + b
Since f(x) is onto, f(-1) = 0
⇒ -a + b = 0 …(1)
& f(1) = 2
⇒ a + b = 2 …(2)
-a + b = 0
Samacheer Kalvi 10th Maths Chapter 1 Relations and Functions Ex 1.4 9

Question 9.
If the function f is defined by
Samacheer Kalvi 10th Maths Chapter 1 Relations and Functions Ex 1.4 10
(i) f(3)
(ii) f(0)
(iii) f(-1.5)
(iv) f(2) + f(-2)
Solution:
(i) f(3) ⇒ f(x) = x + 2 ⇒ 3 + 2 = 5
(ii) f(0) ⇒ 2
(iii) f (- 1.5) = x – 1
= -1.5 – 1 = -2.5
(iv) f(2) + f(-2)
f(2) = 2 + 2 = 4     [∵ f(x) = x + 2]
f(-2) = -2 – 1 = -3    [∵ f(x) = x – 1]
f(2) + f(-2) = 4 – 3 = 1

Question 10.
A function f: [-5,9] → R is defined as follows:
Samacheer Kalvi 10th Maths Chapter 1 Relations and Functions Ex 1.4 11
Solution:
f : [-5, 9] → R
(i) f(-3) + f(2)
f(-3) = 6x + 1 = 6(-3) + 1 = -17
f(2) = 5 × 2 – 1 = 5(22) – 1 = 19
∴ f(-3) + f(2) = -17 + 19 = 2

(ii) f(7) – f(1)
f(7) = 3x – 4 = 3(7) – 4 = 17
f(1) = 6x + 1 = 6(1) + 1 = 7
f(7) – f(1) = 17 – 7 = 10

(iii) 2f(4) + f(8)
f(4) = 5x2 – 1 = 5 × 42 – 1 = 79
f(8) = 3x – 4 = 3 × 8 – 4 = 20
∴ 2f(4) + f(8) = 2 × 79 + 20 = 178
Samacheer Kalvi 10th Maths Chapter 1 Relations and Functions Ex 1.4 12

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Question 11
The distance S an object travels under the influence of gravity in time t seconds is 1 2 given by S(t) = \(\frac { 1 }{ 3 } \)gt2 + at + b, where, (g is the acceleration due to gravity), a, b are constants. Check if the function S(t) is one-one.
Answer:
S(t) = \(\frac { 1 }{ 2 } \)gt2 + at + b
Let the time be 1, 2, 3 …. n seconds
S(1) = \(\frac { 1 }{ 2 } \)g(1)2 + a(1) + b
= \(\frac { g }{ 2 } \) + a + b
S(2) = \(\frac { 1 }{ 2 } \) g(2)2 + a(2) + b
= \(\frac { 4g }{ 2 } \) + 2a + b
= 2g + 2a + b
S(3) = \(\frac { 1 }{ 2 } \) g(3)2 + a(3) + 6
= \(\frac { 9 }{ 2 } \) g + 3a + b
For every different value of t, there will be different distance.
∴ It is a one-one function.

Question 12.
The function ‘t’ which maps temperature in Celsius (C) into temperature in Fahrenheit (F) is defined by t(C)= F where F = \(\frac{9}{5}\) C + 32 . Find,
(i) t(0)
(ii) t(28)
(iii) t(-10)
(iv) the value of C when t(C) = 212
(v) the temperature when the Celsius value is equal to the Farenheit value.
Solution:
Samacheer Kalvi 10th Maths Chapter 1 Relations and Functions Ex 1.4 13
Samacheer Kalvi 10th Maths Chapter 1 Relations and Functions Ex 1.4 14

Samacheer Kalvi 10th Science Solutions Chapter 14 Transportation in Plants and Circulation in Animals

You can Download Samacheer Kalvi 10th Science Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 14 Transportation in Plants and Circulation in Animals

Samacheer Kalvi 10th Science Transportation in Plants and Circulation in Animals Textual Evaluation Solved

I. Choose the Correct Answer

Question 1.
Active transport involves ______.
(a) movement of molecules from lower to higher concentration.
(b) expenditure of energy.
(c) it is an uphill task.
(d) all of the above.
Answer:
(a) movement of molecules from lower to higher concentration.

Question 2.
Water which is absorbed by roots is transported to aerial parts of the plant through:
(a) cortex
(b) epidermis
(c) phloem
(d) xylem
Answer:
(d) xylem

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Question 3.
During transpiration, there is loss of ______.
(a) carbon dioxide
(b) oxygen
(c) water
(d) none of the above.
Answer:
(c) water

Question 4.
Root hairs are:
(a) cortical cell
(b) projection of epidermal cell
(c) unicellular
(d) both (b) and (c)
Answer:
(d) both (b) and (c)

Question 5.
Which of the following process requires energy?
(a) active transport
(b) diffusion
(c) osmosis
(d) all of them.
Answer:
(a) active transport

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Question 6.
The wall of human heart is made of:
(a) Endocardium
(b) Epicardium
(c) Myocardium
(d) All of the above
Answer:
(d) All of the above

Question 7.
Which is the sequence of correct blood flow ______.
(a) ventricle – atrium – vein – arteries
(b) atrium – ventricle – veins – arteries
(c) atrium – ventricle – arteries – vein
(d) ventricles – vein – atrium – arteries.
Answer:
(c) atrium – ventricle – arteries – vein

Question 8.
A patient with blood group 0 was injured in an accident and has blood loss. Which blood group the doctor should effectively use for transfusion in this condition?
(a) O group
(b) AB group
(c) A or B group
(d) all blood group
Answer:
(a) O group

Question 9.
‘Heart of heart’ is called ______.
(a) SA node
(b) AV node
(c) Purkinje fibres
(d) Bundle of His.
Answer:
(a) SA node

Question 10.
Which one of the following regarding blood composition is correct?
(a) Plasma – Blood + Lymphocyte
(b) Serum – Blood + Fibrinogen
(c) Lymph – Plasma + RBC + WBC
(d) Blood – Plasma + RBC + WBC + Platelets
Answer:
(d) Blood – Plasma + RBC + WBC + Platelets

II. Fill in the Blanks

Question 1.
______ involves evaporative loss of water from aerial parts.
Answer:
Transpiration.

Question 2.
Water enters the root cell through a ______ plasma membrane.
Answer:
Osmosis.

Question 3.
Structures in roots that help to absorb water are ______.
Answer:
Root hairs.

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Question 4.
Normal blood pressure is ______.
Answer:
120 mm / 80 mm Hg.

Question 5.
The normal human heartbeat rate is about _______ time per minute.
Answer:
72 – 75.

III. Match the following

Question 1.

1. Symplastic pathway (a) Leaf
2. Transpiration (b) Plasmodesmata
3. Osmosis (c) Pressure in xylem
4. Root Pressure (d) Pressure gradient

Answer:

  1. (b) Plasmodesmata
  2. (a) Leaf
  3. (d) Pressure gradient
  4. (b) Pressure in xylem.

Question 2.

1. Leukaemia (a) Thrombocytes
2. Platelets (b) Phagocyte
3. Monocytes (c) Decrease in leucocytes
4. Leucopenia (d) Blood Cancer
5. AB blood group (e) Allergic condition
6. O blood group (f) Inflammation
7. Eosinophil (g) Absence of antigen
8. Neutrophils (h) Absence of antibody

Answer:

  1. (d) Blood Cancer
  2. (a) Thrombocytes
  3. (b) Phagocyte
  4. (c) Decrease in leucocytes
  5. (h) Absence of antibody
  6. (g) Absence of antigen
  7. (e) Allergic condition
  8. (f) Inflammation.

IV. State whether True or False. If false write the correct statement.

Question 1.
The phloem is responsible for the translocation of food.
Answer:
True.

Question 2.
Plants lose water by the process of transpiration.
Answer:
True.

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Question 3.
The form of sugar transported through the phloem is glucose.
Answer:
False.
Correct Statement: The form of sugar transported through the phloem is Sucrose.

Question 4.
In the apoplastic movement, the water travels through the cell membrane and enter the cell.
Answer:
False.
Correct Statement: In the apoplastic movement, the water travels through the intercellular spaces and walls of the cell.

Question 5.
When guard cells lose water the stoma opens.
Answer:
False.
Correct Statement: When guard cells lose water, the stoma closed.

Question 6.
initiation and stimulation of heartbeat take place by nerves.
Answer:
True.

Question 7.
All veins carry deoxygenated blood.
Answer:
False.
Correct Statement: All veins carry deoxygenated blood except the Pulmonary vein.

Question 8.
WBC defend the body from bacterial and viral infections.
Answer:
True.

Question 9.
The closure of the mitral and tricuspid valves at the start of the ventricular systole produces the first sound ‘LUBB’.
Answer:
True.

V. Answer in a word or Sentence.

Question 1.
Name two-layered protective covering of the human heart.
Answer:
Pericardium.

Question 2.
What is the shape of RBC in human blood?
Answer:
RBC’s are bioconcave and disc shaped.

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Question 3.
Why is the colour of the blood-red?
Answer:
The colour of the blood is red, due to the presence of respiratory pigment Haemoglobin.

Question 4.
Which kind of cells are found in the lymph?
Answer:
Cells found in the lymphatics are lymphocytes.

Question 5.
Name the heart valve associated with the major arteries leaving the ventricles.
Answer:
Semilunar Valves.

Question 6.
Mention the artery which supplies blood to the heart muscle.
Answer:
Heart muscle receive oxygenated blood from coronary arteries that orginate from the aortic arch.

VI. Short Answer Questions.

Question 1.
What causes the opening and closing of guard cells of stomata during transpiration?
Answer:
During transpiration, the movement of (Potassium) ions, in and out of the guard cells, causes the opening and closing of stomate. When the water moves inside the guard cells, causing them to swell up and become turgid, making the stomata open. When guard cells cause water to move out of the cell, make guard cells shrunk, and the stomata pore closes.

Question 2.
What is cohesion?
Answer:
The force of attraction between the water molecules is called cohesion.

Question 3.
Trace the pathway followed by water molecules from the time it enters a plant root to the time it escapes into the atmosphere from a leaf.
Answer:
Samacheer Kalvi 10th Science Solutions Chapter 14 Transportation in Plants and Circulation in Animals 1

Question 4.
What would happen to the leaves of a plant that transpires more water than its absorption in the roots?
Answer:
If the rate of transpiration exceeds the rate of absorption, the stomata get closed the cells lose their turgidity and the plant wilts.

Question 5.
Describe the structure and working of the human heart.
Answer:
The human heart is four-chambered. The two upper thin-walled chambers, Auricle or Atria and the lower thick-walled chambers ventricles. The chambers are separated by a septum. The two auricles are separated by an interatrial septum and the two ventricles are separated by the interventricular septum.

The right atrium receives deoxygenated blood from different parts of the body through superior vena cava, inferior vena cava and coronary sinus. Pulmonary veins bring oxygenated blood from the lungs to the left atrium. Both the right and the left auricles pump blood into the right and left ventricles respectively.

From the right ventricle, the pulmonary arteries supply deoxygenated blood to the lungs. From the left ventricle, the Aorta carries the Oxygenated blood to the various organs of the body. The coronary arteries supply blood to the heart. This process is repeated again and again.

Question 6.
Why is the circulation in man referred to as double circulation?
Answer:
In human, blood passes twice through the heart to supply once to the body.
Double circulation involves: (i) Systemic circulation, (ii) Pulmonary circulation.
(i) Systemic circulation: In systemic circulation, from the left ventricle blood is pumped into the aorta and to various parts of the body.
(ii) Pulmonary circulation: In pulmonary circulation, from right ventricle deoxygenated blood is pumped into pulmonary artery which carries blood to the lungs for oxygenation.

Question 7.
What are heart sounds? How are they produced?
Answer:
The rhythmic closure and opening of the valves cause the sound of the heart. When the closure of the tricuspid and bicuspid valves after the beginning of ventricular systole, the sound ‘LUBB’ is produced. When the closure of the semilunar valve at the end of ventricular systole, the sound ‘DUPP’ is produced.

Question 8.
What is the importance of valves in the heart?
Answer:
The valves in heart are muscular flap that regulates the flow of blood in a single direction and prevents back flow of blood.

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Question 9.
Who discovered the Rh factor? Why was it named so?
Answer:
The Rh factor was discovered by Landsteiner and Wiener in 1940 in Rhesus monkey. So it was named the Rh factor.

Question 10.
How are arteries and veins structurally different from one another?
Answer:

Artery Vein
1. Distributing vessel 1. Collecting vessel
2. Pink in colour 2. Red in colour
3. Deep location 3. Superficial in location
4. Blood flow with high pressure 4. Blood flow with low pressure
5. Wall of an artery is strong, thick and elastic 5. Wall of a vein is weak, thin and non-elastic
6. All arteries carry oxygenated blood except pulmonary arteries 6. All veins carry deoxygenated blood except pulmonary veins
7. Internal valves are absent 7. Internal valves are present

Question 11.
Why is the Sinoatrial node called the pacemaker of heart?
Answer:
Although impulse is produced by the entire neuro muscular pathway, the frequency of impulse generation is maximum in case of Sino atrial node in comparison to other parts of pathway. Hence it guides the rhythm of heart beat and is called the pacemaker of the heart.

Question 12.
Differentiate between systemic circulation and pulmonary circulation.
Answer:

Systemic Circulation Pulmonary Circulation
1. Circulation of oxygenated blood from the left ventricle of the heart to various organs of the body. 1. Circulation starts in the right ventricle of the heart and reaches the lungs with deoxygenated blood.
2. Return of deoxygenated blood to the right atrium. 2. Pulmonary Artery collects the oxygenated blood from the lungs.
3. Aorta carries oxygenated blood to all the organs of the body. 3. The Oxygenated blood is supplied to the left atrium of the heart by the Pulmonary.

Question 13.
The complete events of the cardiac cycle last for 0.8 sec. What is the timing for each event?
Answer:

  1. Auricular systole – Contraction of auricles = 0.1 sec
  2. Ventricular systole – Contraction of ventricle = 0.3 sec
  3. Ventricular diastole – Relaxation of ventricle = 0.4 sec

VII. Give reasons for the following Statements.

Question 1.
Minerals cannot be passively absorbed by the roots.
Answer:
The minerals cannot be passively absorbed by the roots because

  • The minerals are present in the soil as charged particles (ion) and cannot move across the cell membrane.
  • The concentration of minerals in the soil is lower than the concentration of minerals in the root.

Question 2.
Guard cells are responsible for opening and closing of Stomata.
Answer:
Opening and closing of stomata takes place due to changes in turgor of guard cell. The turgor changes in the guard cells are due to entry and exit of water into and out of the guard cells. During the day, water from the subsidiary cells enter the guard cells making it fully turgid causing the stomata to open. During night time, water from guard cells enters the subsidiary cells makes the guard cells flaccid causing the stomata closes.

Question 3.
The movement of substances in the phloem can be in any direction.
Answer:
The movement of substance in the Phloem can be in any direction because, the food to reach the plant parts like stem, leaves, flower, bud, seeds etc, the movement can be upwards or downwards, that is bidirectional.

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Question 4.
Minerals in the plants are not lost when the leaf falls.
Answer:
Minerals are remobilised from older drying leaves to younger leaves. Elements like phosphorous, sulphur, nitrogen, potassium are easily mobilised, while elements like calcium are not remobilised. This can be seen in decidous leaves.

Question 5.
The walls of the right ventricle are thicker than the right auricles.
Answer:
The walls of the right ventricle are thicker than the right auricle because, the right ventricle has to pump out the blood with force to the Pulmonary trunk, which bifurcates to form the right and left Pulmonary Arteries.

Question 6.
Mature RBC in mammals do not have cell organelles.
Answer:
The RBCs are devoid of nucleus, mitochondria ribosome and endoplasmic reticulum. The absence of these organelles accommodates more haemoglobin thereby maximising the oxygen carrying capacity of the cell. Biconcave shape increase the surface area for oxygen binding, loss of mitochondria allow the RBC to transport all the oxygen to tissues and loss of endoplasmic reticulum allows more flexibility for RBC to move through the narrow capillaries.

VIII. Long Answer Questions.

Question 1.
How do plants absorb water? Explain.
Answer:
There are millions of root hairs on the tip of the root, which absorb water and minerals by diffusion. Diffusion takes place across cell membranes. Root hairs are a thin-walled, slender, extension of Epidermal cell, that increases the surface area of absorption. Active transport utilises energy to pump molecules against a concentration gradient. Active transport is carried out by membrane – bound proteins. These proteins use energy to carry substances across the cell membrane.

The cell wall of root hair is permeable and allows the water and minerals to enter. The cell membrane is semi – permeable. So it allows movement of water molecules from the region of higher concentration to the region of lower concentration. Once the water enters the root hairs, the concentration of water molecules in the root hair cells become more than that of Cortex. So the water from the root hair moves to the cortical cells by osmosis and then reaches the xylem.

Due to transpiration, the water is lost from the leaves and pressure is created at the top to pull more water from the xylem to the mesophyll cells, by the process of Transpiration pull. This extends up to the roots causing the roots to absorb more water from the soil to ensure the continuous flow of water from the roots to the leaves.

Question 2.
What is transpiration? Give the importance of transpiration.
Answer:
The loss of water from the aerial parts of plant in the form of vapours is called transpiration.
Importance of transpiration:

  1. Creates transpirational pull for transport of water.
  2. Supplies water for photosynthesis.
  3. Transports minerals from soil to all parts of the plant.
  4. Cools the surface of the leaves by evaporation.
  5. Keeps the cells turgid; hence, maintains their shape.

Question 3.
Why are leucocytes classified as granulocytes and agranulocytes? Name each cell and mention its functions.
Answer:
White blood corpuscles are colourless. They are nucleated cells. They are found in the bone marrow, spleen, thymus and lymph nodes. They are grouped into two categories:
1. Granulocytes: They contain granules in their cytoplasm. Their nucleus is irregular or lobed. The granulocytes are of three types:

  • Neutrophils: They are large in size and have a 2 – 7 lobed nucleus. Their numbers are increased during infection and inflammation.
  • Eosinophils: It has a bilobed nucleus. Their numbers increases during conditions of allergy and parasitic infections. It brings about the detoxification of toxins.
  • Basophils: Basophils have a lobed nucleus. They release chemicals during the process of inflammation.

2. Agranulocytes: Granules are not found in the cytoplasm of these cells. Thy is of two types:

  • Lymphocytes: Lymphocytes produce antibodies during bacterial and viral infections.
  • Monocytes: They are the largest of the leucocytes and are amoeboid in shape. They are phagocytic and can engulf bacteria.

Question 4.
Differentiate between systole and diastole. Explain the conduction of heart beat.
Answer:
systole:
The contraction of heart is called systole.
diastole:
The relaxation of heart is called diastole.
Conduction of heart beat : The heart in human is myogenic. The cardiac cells with fastest rhythm are called the pacemaker cells. These cells are located in the right sino-atrial node. The impulse from the sino-atrial node. spreads as a wave of contraction over the right and left atrial wall pushing the blood through the atrio ventricular valves into the ventricle. Two special cardiac muscles fibres originate from the auriculo ventricular node and are called the bundle of his which runs down into the interventricular septum and the fibres spread into the ventricles. These fibres are called the Purkinje.

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Question 5.
Enumerate the functions of blood.
Answer:
Functions of blood:

  • Transport of respiratory gases (Oxygen and CO2).
  • Transport of digested food materials to different body cells.
  • Transport of hormones.
  • Transport of nitrogenous excretory products like ammonia, urea and uric acid.
  • It is involved in the protection of the body and defence against diseases.
  • It acts as a buffer and also helps in the regulation of pH and body temperature.
  • It maintains proper water balance in the body.

IX. Assertion and Reasoning Questions.

Direction: In each of the following questions a statement of assertion (A) is given and a corresponding statement of Reason (R) is given just below it. Mark the correct statement as:
(a) If both A and R are true and R is the correct explanation of A.
(b) If both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) Both A and R are false.

Question 1.
Assertion: RBC plays an important role in the transport of respiratory gases.
Reason: RBC does not have cell organelles and nucleus.
Answer:
(a) If both A and R are true and R is the correct explanation of A.

Question 2.
Assertion: Persons with AB blood group are called universal recipients because they can receive blood from all groups.
Reason: Antibodies are absent in persons with AB blood group.
Answer:
(a) If both A and R are true and R is the correct explanation of A.

X. Higher Order Thinking Skills (HOTS) Questions

Question 1.
When any dry plant material is kept in water, they swell up. Name and define the phenomenon involved in this change.
Answer:
The swelling up is due to Imbibition. Imbibition is a type of diffusion in which a solid absorbs water and gets swelled up. If Imbibition were not there, seedlings would not have been able to emerge out of the soil.

Question 2.
Why are the wails of the left ventricle thicker than the other chambers of the heart?
Answer:
The left ventricles have thick walls because the ventricle have to pump out blood with force away from the heart.

Question 3.
Doctors use a stethoscope to hear the sound of the heart. Why?
Answer:
The heart sound is heard by placing the stethoscope on the chest. It is a useful diagnostic tool to identify and localize the health problems and diagnose disease.

Question 4.
How does the pulmonary artery and pulmonary vein differ in their function when compared to a normal artery and vein?
Answer:

  1. All arteries carry oxygenated blood except the pulmonary artery which carry deoxygenated blood to the lungs.
  2. All veins carry deoxygenated blood except the pulmonary vein which carry oxygenated blood from the lungs to the heart.

Question 5.
Transpiration is a necessary evil in plants. Explain.
Answer:
Transpiration is a necessary evil in plants because it is inevitable but potentially harmful. Loss of water from the plant results wilting and cause the death of a plant if a condition of drought is experienced.
But transpiration is a great significance for the plant.

  • Water is conducted, in most tall plants due to transpiration pull.
  • Minerals dissolved in water are distributed throughout the plant body by Transpiration Stream.
  • Evaporation of water from the cells of leaves has a cooling effect on plants.
  • The wet surface of leaf cells allows gaseous exchange.

Textbook Activities Solved

Activity 1.
Demonstration of Osmosis: A thistle funnel whose mouth is covered with a semipermeable membrane is filled with sucrose solution. It is kept inverted in a beaker containing water. The water will diffuse across the membrane due to osmosis and raise the level of the solution in the funnel.
Answer:
A thistle funnel, whose mouth is covered with parchment paper, which acts as a semi-permeable membrane, is filled with sucrose solution. The funnel is kept inverted in a beaker, containing water. The water from the beaker enters or diffuse across the membrane from the region of water higher concentration to the water lower concentration (Sucrose Solution) through Semi – Permeable membrane, by osmosis. This causes the rise of the level of a solution in the funnel.
Samacheer Kalvi 10th Science Solutions Chapter 14 Transportation in Plants and Circulation in Animals 2

Activity 2.
Demonstration of Root Pressure: Choose a small soft – stemmed plant. Cut the stem horizontally near the base with a blade in the morning. You will see drops of solution oozing out of the cut stem due to root pressure.
Answer:
We will see drops of solution Oozing out of the cell stem. The Root Pressure, which pushes ^the solution up from the base.

Activity 3.
Determining Heart Rate?
Answer:
Materials:

  1. Stopwatch or Stop clock.

Procedure:

  1. Have your partner to find the pulse in your wrist and count your heartbeats for 15 seconds while you are seated. Calculate your resting heart rate in beats per minute.
  2. Have your partner to count your heart beats for 15 seconds after you jog or run for 5 minutes. Calculate your heart rate in beats per minute.

Analyse:
(i) What causes your pulse?
(ii) What causes the change in your heartbeat rate in each situation? You can write the answer yourself.
Answer:
(i) Pulse, rhythmic dilation of an artery generated by the opening and closing of the aortic valve in the heart. A pulse can be felt by applying firm fingertip pressure to the skin at sites where the arteries travel near the skin’s surface; it is more evident when surrounding muscles are relaxed.
(ii) You can write the answer yourself.

Samacheer Kalvi 10th Science Transportation in Plants and Circulation in Animals Additional Questions Solved

I. Fill in the blanks.

Question 1.
______ is responsible for the movement of water up to the base of the stem.
Answer:
Root pressure.

Question 2.
The other name for Red blood corpuscles (RBC) is called ______.
Answer:
Erythrocytes.

Question 3.
The apices of the flaps of tricuspid valves are held in position by ______.
Answer:
Chordaetendinae.

Question 4.
The supply of blood to the heart muscles is called ______ circulation.
Answer:
Coronary.

Question 5.
Sinuses are the body cavities which are called ______.
Answer:
Haemocoel.

II. Write True or False for the following statements. Write the true statement for the false statement.

Question 1.
Root hairs are a thin-walled, slender extension of Parenchyma cells.
Answer:
False.
Correct Statement: Root hairs are a thin-walled, slender extension of epidermal cell.

Question 2.
The transpiration pull sucks the water column from the xylem tubes so that the water is able to rise up in the tallest plants. Answer:
True.

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Question 3.
The force of attraction between the molecules of companion cells is called cohesion.
Answer:
False.
Correct Statement: The force of attraction between the molecules of water is called cohesion,

Question 4.
Blood platelets help in the clotting of blood, form clot at the site of injury and prevent blood loss.
Answer:
True.

III. Match the following.

Question 1.

1. Universal Recipient (a) Left Atrio Ventricular Valve
2. Blood (b) Thin and non-elastic cells
3. Thymocytes (c) Blood group AB
4. Universal Donor (d) Thick and elastic cells
5. Mitral (e) Blood group ‘O’
6. Veins (f) Blood platelets
7. Arteries (g) Connective tissue

Answer:

  1. (c) Blood group AB
  2. (g) Connective tissue
  3. (f) Blood platelets
  4. (e) Blood group ‘O’
  5. (a) Left Atrio Ventricular Valve
  6. (b) Thin and non-elastic cells
  7. (d) Thick and elastic cells

IV. Answer the following in a word or a Sentence.

Question 1.
What is Root Pressure?
Answer:
As ion from the Soil is actively transported into the Vascular tissue of the root, water moves along and increases the pressure inside the Xylem. This pressure is called root pressure.

Question 2.
What is the role of valves in heart?
Answer:
The valves are muscular flap that regulates the flow of blood in a single direction and prevent back flow of blood.

Question 3.
What is capillary action?
Answer:
Water or any liquid rises in a capillary tube because of physical forces. This phenomenon is called capillary action.

Question 4.
What is normal heart beat in man? How does it occur?
Answer:
The normal heart beat in man is about 72 – 75 times per minute. Rhythmic contraction and expansion of heart causes heart beat.

Question 5.
What are capillaries?
Answer:
Capillaries are narrow tubes formed by branching of arterioles which then unite to form the venules and veins. Capillaries are formed of a single layer of endothelial cells.

Question 6.
What is a pulse?
Answer:
The expansion of the artery every time, the blood is forced into it is called pulse. Normal pulse rate ranges from 70 – 90 per minute.

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Question 7.
What is Cardiac Cycle?
Answer:
The sequence of events occurring from the beginning to the completion of one heartbeat is called the Cardiac Cycle.

Question 8.
What is hypertension?
Answer:
The condition, where a prolonged or constant elevation of blood pressure exist is called hypertension (High blood pressure)

Question 9.
What is Apoplast pathway?
Answer:
The apoplastic movement of water occurs through the intercellular spaces and the walls of the cells with gradient energy.

Question 10.
Why are membrane – bound proteins called pumps?
Answer:
Active transport is carried out by membrane-bound proteins. These proteins use energy to carry substances across the cell membrane. So they are called pumps.

Question 11.
When does plasmolysis occur?
Answer:
Plasmolysis occur when water moves out of the cell and resulting in the shrinkage of cell membrane away from the cell wall.

Question 12.
What are Lymph nodes?
Answer:
Lymph nodes are small oval or fan-shaped structures, located along the length of lymphatic vessels.

V. Answer the following briefly.

Question 1.
Define the following.

  1. Diffusion
  2. Osmosis
  3. Plasmolysis

Answer:

  1. Diffusion: The movement of molecules in liquid and solids from a region of higher concentration to a region of lower concentration, without the utilization of energy is called Diffusion.
  2. Osmosis: Osmosis is the movement of solvent or water molecules, from the region of higher concentration to the region of lower concentration through a semi – permeable membrane.
  3. Plasmolysis: Plasmolysis is a process, which occurs, when water moves out of the cell and resulting in the shrinkage of the cell membrane, away from the cell wall.

Question 2.
Explain cardiac cycle.
Answer:
The sequence of events occurring from the beginning to the completion of one heart beat is called cardiac cycle. During cardiac cycle blood flows ’ through the chambers of the heart in a specific direction. Each cardiac cycle lasts about 0.8 second. The events during a single cardiac cycle involves

  1. Atrial systole: Contraction of auricles (0.1 sec)
  2. Ventricular systole: Contraction of ventricles (0.3 sec)
  3. Ventricular diastole: Relaxation of ventricles (0.4 sec)

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Question 3.
Explain the two types of Circulatory System.
Answer:
The two types of Circulatory System in animals are

  1. Open type: The blood is pumped by the heart into blood vessels, that open into blood spaces called sinuses, which are the body cavities called haemocoel. Capillary System is absent. Eg. Arthropods, Molluscs, Ascidians.
  2. Clsed type: The blood flows in a complete circuit around the body, through specific blood vessels. The blood flows from arteries to veins, through small blood vessels called capillaries. Eg. Vertebrates.

Question 4.
What is Blood Pressure? How is blood pressure expressed?
Answer:
The force exerted during the flow of blood against the lateral walls of arteries is called Blood Pressure. Blood pressure is usually expressed in terms of systolic pressure and diastolic pressure.

  1. Systolic pressure: The pressure, during ventricular systole, the left ventricle contracts and forces the blood into the aorta is called systolic pressure.
  2. Diastolic pressure: The pressure, during diastole, the ventricle relaxes and the pressure falls to the lowest value, is called Diastolic pressure.

In a healthy adult, during the normal resting condition, the systolic and diastolic blood pressure is expressed as 120mm / 80mm Hg.

Question 5.
In a Tabular Column, mention the distribution of Antigen (RBC) and antibody (Plasma) in different blood groups.
Answer:

Blood Group Antigens on RBC Antibodies in Plasma Can donate to Can receive from
A Antigen A anti – b A and AB A and O
B Antigen B anti – a B and AB B and O
AB Antigen A and B No Antibody AB A, B, AB and O (Universal Recipient)
O No Antigen Both anti a and b A, B, AB and O (Universal Donor) O

Question 6.
Write a note on valves of human.
Answer:
Valves : The valves are the muscular flaps that regulate the flow of blood in a single direction and prevent back flow of blood. The heart contains three types of valves. Right atrioventricular valve : It is located between the right auricle and right ventricle.

It has three thin triangular leaf like flaps and therefore called tricuspid valve. The apices of the flaps are held in position by chordae tendinae arising from the muscular projection of the ventricle wall known as papillary muscles.

Left atrioventricular valve : It is located between the left auricle and left ventricle. It has two cusps and therefore called bicuspid or mitral valve.
Semilunar valves : The major arteries (pulmonary artery and aorta) which leave the heart have semilunar valves which prevent backward flow of blood into the ventricles. They are the pulmonary and aortic semilunar valves.

Question 7.
What is Sphygmo Manometer? Name the instruments used to measure Blood pressure.
Answer:
The sphygmomanometer is a clinical instrument used to measure blood pressure when a person is in a relaxed and resting condition. The pressure of the brachial artery is measured. It helps to estimate the state of blood circulation and the working of the heart. It helps to diagnose conditions such as increased or decreased blood pressure. Monometric and modem digital types are used to measure blood pressure.

Question 8.
What are the factors which affect transpiration?
Answer:
Transpiration is affected by several external and internal factors.

  • External factors: Temperature, light, humidity, and wind speed.
  • Internal factors: Number and distribution of Stomata, Percentage of open stomata, Water status of the plant and Canopy Structure.

VI. Draw a labelled diagram for the following.

Question 1.
(a) Root tip with root hairs.
(b) Guard cell in turgid and Flacid condition.
Answer:
(a)
Samacheer Kalvi 10th Science Solutions Chapter 14 Transportation in Plants and Circulation in Animals 3
(b)
Samacheer Kalvi 10th Science Solutions Chapter 14 Transportation in Plants and Circulation in Animals 4
Samacheer Kalvi 10th Science Solutions Chapter 14 Transportation in Plants and Circulation in Animals 5

Question 2.
Draw a labelled diagram of the internal structure of Human Heart.
Answer:
Samacheer Kalvi 10th Science Solutions Chapter 14 Transportation in Plants and Circulation in Animals 6

VII. Answer the following in detail.

Question 1.
Explain the types of Blood Circulation.
Answer:
The blood circulates in our body as oxygenated and deoxygenated blood. The types of circulation are:

  1. Systemic Circulation: Circulation of oxygenated blood from the left ventricle of the heart to various organs of the body and return of deoxygenated blood to the right atrium. Aorta carries oxygenated blood to all the organs of the body.
  2. Pulmonary Circulation: The path of pulmonary circulation starts in the right ventricle. Pulmonary artery arises from the right ventricle and reaches the lungs with deoxygenated blood. Pulmonary veins collect the oxygenated blood from the lungs and supplies it to the left atrium of the heart.
  3. Coronary circulation: The supply of blood to the heart muscles (cardiac muscles) is called coronary circulation. Cardiac muscles receive oxygenated blood from coronary arteries, that originate from the aortic arch. Deoxygenated blood from the cardiac muscles, drains into the right atrium by the coronary sinuses.

In human, the blood circulates twice, through the heart in one complete cycle, called double circulation. The oxygenated blood does not mix with the deoxygenated blood.

Question 2.
Explain the Lymphatic System with diagram and mention the function of Lymph.
Answer:
The lymphatic system consists of lymphatic capillaries, lymphatic vessels, lymph nodes and lymphatic ducts.
Lymph is a fluid, that flows through the lymphatic system. The lymphatic capillaries unite to form large lymphatic vessels. Lymph nodes are small, oval or pear-shaped structures located along the lymphatic vessels.
Samacheer Kalvi 10th Science Solutions Chapter 14 Transportation in Plants and Circulation in Animals 7
Lymph from the intercellular spaces drains into lymphatic capillaries. Lymph is a colourless fluid formed when Plasma, Proteins and blood cells escape into intercellular spaces in the tissues, through the pores present in the walls of the capillaries. It is similar to blood plasma, but is colourless and contains fewer proteins. The lymph contains a very small amount of nutrients, Oxygen, CO2, water and WBC.

Functions of Lymph:

  • Supplies nutrients and oxygen to those parts where blood cannot reach.
  • It drains away excess tissue fluid and metabolites and returns proteins to the blood from tissue spaces.
  • The lymph also carries absorbed fats from small intestine to the blood. The lymphatic capillaries of internal villi (lacteals) absorb digested fats.
  • Lymphocytes in the lymph defend the body from infections.

VIII. Higher Order Thinking Skills (HOTS) Questions

Question 1.
What do you know about the following?

  1. Anaemia
  2. Leucocytosis
  3. Leukopenia
  4. Thrombocytopenia

Answer:

  1. Anaemia → Decrease in the number of Erythrocytes.
  2. Leucocytosis → Increase in the number of Leukocytes.
  3. Leukopenia → Decrease in the number of Leukocytes.
  4. Thrombocytopenia → Decrease in the number of Thrombocytes.

Question 2.
The cardiac pacemaker in a patient fails to function normally. The doctor finds that an artificial pacemaker is to be grafted in him. It is likely that it will be drafted at the site of.
Answer:
Sino-atrial node, which acts as the pacemaker of the heart because it is capable of initiating impulse which can stimulate the heart muscles to contraction.

Question 3.
What do you know about the following important terms, which we come across every day?

  1. Heart attack
  2. Cardiac arrest
  3. Heart Failure.

Answer:

  1. Heart attack: The blood flow to the heart is blocked. A blockage of blood flow to the heart muscle. The heart muscle affected from lack of blood supply.
  2. Cardiac arrest: The heart stops beating and needs to be restarted. Cardiac arrest is an electrical problem, triggered by a disruption of the heart’s rhythm.
  3. Heart Failure: Heart Failure occurs, when the heart muscle fails to pump as much as blood as the body needs. It may come to a person suddenly.

Question 4.
Removal of ring wood of tissue outside the vascular cambium from the tree trunk kills it because:
Answer:
Food does not travel down and root becomes starved (Translocation of food).

Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.1

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.1

Question 1.
Solve the following system of linear equations in three variables
(i) x + y + z = 5; 2x – y + z = 9; x – 2y + 3z = 16
(ii) \(\frac { 1 }{ x } \) – \(\frac { 2 }{ y } \) + 4 = 0; \(\frac { 1 }{ y } \) – \(\frac { 1 }{ z } \) + 1 = 0; \(\frac { 2 }{ z } \) + \(\frac { 3 }{ x } \) = 14
(iii) x + 20 = \(\frac { 3y }{ 2 } \) + 10 = 2z + 5 = 110 – (y + z)
Solutions:
(i) x + y + z = 5 ………….. (1)
2x – y + z = 9 …………. (2)
x – 2y + 3z = 16 …………. (3)
Samacheer Kalvi 10th Maths Chapter 3 Algebra Ex 3.1 1
Substitute z = 4 in (4)
3x + 2(4) = 14
3x + 8 = 14
3x = 6
x = 2
Substitute x = 2, z = 4 in (1)
2 + y + 4 = 5 ⇒ y = -1
x = 2, y = -1, z = 4

You can Download Samacheer Kalvi 10th Maths Solution Book Pdf Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Samacheer Kalvi 10th Maths Chapter 3 Algebra Ex 3.1 2

\(\frac{1}{y}\) = b
\(\frac{1}{z}\) = c in (1), (2) & (3)
a – 2b + 4 = 0 ⇒ a – 2b = -4 …………. (1)
b – c + 1 = 0 ⇒ b – c = -1 ……….. (2)
2c + 3a = 14 ⇒ 2c + 3a = 14 …………. (3)
Samacheer Kalvi 10th Maths Chapter 3 Algebra Ex 3.1 3

(iii) x + 20 = \(\frac { 3y }{ 2 } \) + 10 = 2z + 5 = 110 – (y + z)
x = \(\frac { 3y }{ 2 } \) – 10 …………. (1)
2z + 5 = 110 – (y + z)
2z = 105 – y – z
y = 105 – 3z ………….. (2)
Substitute (2) in (1), x = \(\frac { 315 }{ 2 } \) – \(\frac { 9z }{ 2 } \) – 10
= 2z + 5 – 20
∴ 315 – 9z – 20 = 4z – 30
13 z = 315 – 20 + 30
= 325
z = \(\frac { 325 }{ 13 } \) = 25
x + 20 = 2z + 5
x + 20 = 50 + 5
x = 35
Substitute z = 25 in (2)
y = 105 – 3z = 105 – 75 = 30
y = 30
x = 35, y = 30, z = 25
The system has unique solutions.

Question 2.
Discuss the nature of solutions of the following system of equations
(i) x + 2y – z = 6 ; -3x – 2y + 5z = -12 ; x – 2z = 3
(ii) 2y + z = 3 (-x + 1); -x + 3y -z = -4 3x + 2y + z = – \(\frac { 1 }{ 2 } \)
(iii) \(\frac { y+z }{ 4 } \) = \(\frac { z+x }{ 3 } \) = \(\frac { x+y }{ 2 } \); x + y + z = 27
Solution:
(i) x + 2y – z = 6 …………. (1)
-3x – 2y + 5z = -12 ……… (2)
x – 2z = 3 …………… (3)
Samacheer Kalvi 10th Maths Chapter 3 Algebra Ex 3.1 4
We see that the system has an infinite number of solutions.
(ii) 2y + z = 3(-x + 1);
-x + 3y – z = -4;
3x + 2y + z = –\(\frac { 1 }{ 2 } \)
2y + z + 3x = 3 ⇒ 3x + 2y + z = 3 ………….. (1)
-x + 3y – z = -4 …………. (2)
3x + 2y + z = –\(\frac { 1 }{ 2 } \) ………………. (3)
Samacheer Kalvi 10th Maths Chapter 3 Algebra Ex 3.1 5
Samacheer Kalvi 10th Maths Chapter 3 Algebra Ex 3.1 6
This is a contradiction. This means the system is inconsistent and has no solutions.
Samacheer Kalvi 10th Maths Chapter 3 Algebra Ex 3.1 7
Sub. x = 3 in (4) ⇒ 5(3) – z = 0
15 – z = 0
-z = -15
z = 15
Sub, x = 3, z = 15 in (3)
x + y + z = 27
3 + y + 15 = 27
y = 27 – 18 = 9
x = 3, y = 9, z = 15
∴ The system has unique solutions.

Question 3.
Vani, her father and her grand father have an average age of 53. One-half of her grand father’s age plus one-third of her father’s age plus one fourth of Vani’s age is 65. Four years ago if Vani’s grandfather was four times as old as Vani then how old are they all now?
Solution:
Let Vani’s age be x
Let Vani’s father’s age be y
Let Vani’s grand father’s age be z.
Samacheer Kalvi 10th Maths Chapter 3 Algebra Ex 3.1 8
Sub, z = 84 in (3), we get
4x – 84 = 12
4x = 96
x = 24
Sub, x = 24, z = 84 in (1) we get
24 + y + 84 = 159
y = 159 – 108
= 51
∴ Vani’s age = 24 years
Her father’s age =51 years
Her grand father’s age = 84 years.

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Question 4.
The sum of the digits of a three-digit number is 11. If the digits are reversed, the new number is 46 more than five times the former number. If the hundreds digit plus twice the tens digit is equal to the units digit, then find the original three digit number?
Solution:
Let the number be 100x + 10y + z.
Reversed number be 100z + 10y + x.
x + y + z = 11 …………… (1)
100z + 10y + x = 5(100x + 10y + z) + 46
100z + 10y + x = 500x + 50y + 5z + 46
499x + 40y – 95z -46 ………….. (2)
x + 2y = z
x + 2y – z = 0 ……………. (3)
Samacheer Kalvi 10th Maths Chapter 3 Algebra Ex 3.1 9

Question 5.
There are 12 pieces of five, ten and twenty rupee currencies whose total value is ₹105. When first 2 sorts are interchanged in their numbers its value will be increased by ₹20. Find the number of currencies in each sort.
Solution:
Let x, y and z be number of currency pieces of 5,10,20 rupees
x + y + z = 12 ………. (1)
5x + 10y + 20z = 105 ………… (2)
10x + 5y + 20z = 125 …………. (3)
Samacheer Kalvi 10th Maths Chapter 3 Algebra Ex 3.1 10
Sub, z = 2 in (5), we get
15y + 20 × 2 = 85
15y = 45
y = 3
Sub; y = 3, z = 2 in (1)
x + y + z = 12
x = 7
∴ The solutions are
the number of ₹ 5 are 7
the number of ₹ 10 are 3
the number of ₹ 20 are 2

Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.1

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.1

Question 1.
Find the area of the triangle formed by the points
(i) (1, -1), (-4, 6) and (-3, -5)
(ii) (-10, -4), (-8, -1) and (-3, -5)
Solution:
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 1
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 2

(ii) (-10, -4), (-8, -1) and (-3, -5)
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 3
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 4

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Question 2.
Detemine whether the sets of points are collinear ?
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 5
Solution:
(i)
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 6
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 7
∴ The given points are collinear
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Question 3.
Vertices of given triangles are taken in order and their areas are provided aside. In each case, find the value of ‘p’.
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 9
Solution:
Area 20 sq. units.
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 10
8p = 104
p = 13

Question 4.
In each of the following, find the value of ‘a’ for which the given points are collinear.
(i) (2, 3), (4, a) and (6, -3)
(ii) (a, 2 – 2a), (-a + 1, 2a) and(-4 – a,6 – 2a)
Solution:
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 11
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 12
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 13

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Question 5.
Find the area of the quadrilateral whose vertices are at
(i) (-9, -2), (-8, -4), (2, 2) and (1, -3)
(ii) (-9, 0), (-8, 6), (-1, -2) and (-6, -3)
Solution:
(i) (-9, -2), (-8, -4), (2, 2), and (1, -3)
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 14
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 15

(ii) (-9, 0), (-8, 6), (-1, -2) and (-6, -3)
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 16
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 17

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Question 6.
Find the value of k, if the area ofa quadrilateral is 28 sq.units, whose vertices are (-4, -2), (-3, k), (3, -2) and (2, 3)
Solution:
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 18
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 19

Question 7.
If the points A(-3, 9) , B(a, b) and C(4,-5) are collinear and if a + b = 1, then find a and b.
Solution:
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 20
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 21

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Question 8.
Let P(11, 7), Q(13.5, 4) and R(9.5, 4) be the mid-points of the sides AB, BC and AC respectively of ∆ABC . Find the coordinates of the vertices A, B and C. Hence find the area of ∆ABC and compare this with area of ∆PQR.
Solution:
p (11, 7), Q (13.5, 4), and R (9.5, 4) are the mid points of the sides of a ∆ABC.
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 22
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 23
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Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 27
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 70

Question 9.
In the figure, the quadrilateral swimming pool shown is surrounded by concrete patio. Find the area of the patio.
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 50
Solution:
Area of the patio = Area of the quadrilateral ABCD – Area of the swimming pool EFGFI.
Area of the quadrilateral ABCD
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 60

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Question 10.
A triangular shaped glass with vertices at A(-5, -4), B(1, 6) and C(7, -4) has to be painted.
If one bucket of paint covers 6 square feet, how many buckets of paint will be required to paint the whole glass, if only one coat of paint is applied.
Solution:
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 90
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 91

Question 11.
In the figure, find the area of
(i) triangle AGF
(ii) triangle FED
(iii) quadrilateral BCEG
Solution:
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 92
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 93
Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Ex 5.1 94

Samacheer Kalvi 10th Science Solutions Chapter 13 Structural Organisation of Animals

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Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 13 Structural Organisation of Animals

Samacheer Kalvi 10th Science Structural Organisation of Animals Textual Evaluation Solved

I. Choose the correct answer.

Question 1.
In leech, locomotion is performed by _____.
(a) Anterior sucker
(b) Posterior sucker
(c) Setae
(d) None of the above.
Answer:
(d) None of the above.

Question 2.
The segments of leech are known as:
(a) Metameres (somites)
(b) Proglottids
(c) Strobila
(d) All the above
Answer:
(a) Metameres (somites)

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Question 3.
Pharyngeal ganglion in leech is a part of _____.
(a) Excretory system
(b) Nervous system
(c) Reproductive system
(d) Respiratory system.
Answer:
(b) Nervous system

Question 4.
The brain of leech lies above the:
(a) Mouth
(b) Buccal Cavity
(c) Pharynx
(d) Crop
Answer:
(c) Pharynx

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Question 5.
The body of leech has _____.
(a) 23 segments
(b) 33 segments
(c) 38 segments
(d) 30 segments.
Answer:
(b) 33 segments

Question 6.
Mammals are ______ animals.
(a) Cold – blooded
(b) Warm – blooded
(c) Poikilothermic
(d) All the above.
Answer:
(b) Warm – blooded

Question 7.
The animals which give birth to young ones are:
(a) Oviparous
(b) Viviparous
(c) Ovoviviparous
(d) All the above
Answer:
(b) Viviparous

II. Fill in the blanks.

Question 1.
The posterior sucker is formed by the fusion of the _______ segments.
Answer:
Last seven.

Question 2.
The existence of two sets of teeth in the life of an animal is called ______ dentition.
Answer:
Diphyodont.

Question 3.
The anterior end of leech has a lobe-like structure called _____.
Answer:
Sucker.

Question 4.
The blood-sucking habit of a leech is known as _____.
Answer:
Sanguivorous.

Question 5.
______ separate nitrogenous waste from the blood in the rabbit.
Answer:
Nephrons.

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Question 6.
_____ spinal nerves are present in the rabbit.
Answer:
37 pairs of.

III. Identify whether the statements are true or false. Correct the false statement.

Question 1.
An anticoagulant present in the saliva of the leech is called heparin.
Answer:
False.
Correct Statement: The anticoagulant present in the saliva of the leech is called Hirudin.

Question 2.
The vas deferens serves to transport the ovum.
Answer:
False.
Correct Statement: The vas deferens serves to transport the sperm.

Question 3.
The rabbit has a third eyelid called tympanic membrane which is movable.
Answer:
False.
Correct Statement: The rabbit has a third eyelid called Nictitating membrane, which is movable.

Question 4.
Diastema is a gap between premolar and molar teeth in rabbit.
Answer:
True.

Question 5.
The cerebral hemispheres of the rabbit are connected by a band of nerve tissue called corpora quadrigemina.
Answer:
False.
Correct Statement: The cerebral hemisphere of the rabbit are connected by a band of nerve tissue called Corpus callosum.

IV. Match the Columns I, II and III correctly.

Question 1.

Organs Membranous Covering Location
Brain Pleura abdominal cavity
Kidney Capsule Mediastinum
Heart Meninges enclosed in the thoracic cavity
Lungs Pericardium cranial cavity

Answer:

Organs Membranous Covering Location
Brain Meninges cranial cavity
Kidney Capsule abdominal cavity
Heart Pericardium enclosed in the thoracic cavity
Lungs Pleura mediastinum

V. Answer in a sentence.

Question 1.
Give the common name of the Hirudinaria granulosa.
Answer:
Indian Cattle Leech.

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Question 2.
How does leech respire?
Answer:
The leech respire through skin (diffusion).

Question 3.
Write the dental formula of the rabbit.
Answer:
Dental formula is \(\left(\mathrm{I} \frac{2}{1}, \mathrm{C} \frac{0}{0}, \mathrm{PM} \frac{3}{2}, \mathrm{M} \frac{3}{3}\right)\) in Rabbit, which is written as \(\frac{2033}{1023}\)

Question 4.
How many pairs of testes are present in leech?
Answer:
In leech, there are eleven pairs of testes. One pair in each segment from 12 to 22 segments.

Question 5.
How is diastema formed in rabbit?
Answer:
The diastema is formed in Rabbit, as a gap between the incisors and premolars,

Question 6.
What organs are attached to the two bronchi?
Answer:
Lung, Bronchioles and Alveoli are the Organs, attached to Bronchi.

Question 7.
Which organ acts as suction pump in leech?
Answer:
The muscular pharynx found below the buccal cavity acts as suction pump in leech. It is surrounded by large masses, unicellular glands called salivary glands.

Question 8.
What does CNS stand for?
Answer:
CNS stands for Central Nervous System.

Question 9.
Why are the teeth of a rabbit called heterodont?
Answer:
The teeth of Rabbit are of different types. So it is called heterodont.

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Question 10.
How does leech suck blood from the host?
Answer:
In leech, blood is sucked by the muscular pharynx. The salivary secretion is poured in the wound. The saliva contains an active substance called hirudin which prevents the coagulation of the blood.

VI. Short Answer Questions.

Question 1.
Why are the rings of cartilages found in the trachea of a rabbit?
Answer:
Tracheal walls are supported by rings of cartilage, which help in the free passage of air.

Question 2.
List out the parasitic adaptations in leech.
Answer:

  1. The suckers are the primary organ of the blood sucking.
  2. The blood is sucked by muscular pharynx.
  3. Leeches attaches itself to the body of host by Anterior and Posterior ends of the body.
  4. The three jaws inside the mouth, causes a painless triradiate or Y shaped incision in the skin of the host.
  5. A protein called hirudin is produced in the salivary gland of leech to prevent blood coagulation. Thus, a continuous supply of the blood is maintained.
  6. Parapodia and setae are completely absent.
  7. Leeches also inject an anaesthetic substances that prevents the host from feeling their bite.
  8. In the crop, blood is stored which gives nourishment to the leech for several months. Due to this reason there is no eloborate secretion of the digestive juices and enzymes.

VII. Long Answer Questions.

Question 1.
How is the circulatory system designed in leech to compensate the heart structure?
Answer:
There are no true blood vessels in leeches. The blood vessels are replaced by canals called haemocoelic canals. These canals are filled with haemocoelic fluid. There are four longitudinal canals. One is dorsal lying above the alimentary canal, another is ventral lying below the alimentary canal. The remaining two are lateral lying on either side of the alimentary canal. These four canals are connected together at the posterior end. There is no heart, but the lateral channels serve as a heart by being contractile. They have values inside. The dorsal and ventral channels are non-contractile having no muscular walls.

Question 2.
How does locomotion take place in leech?
Answer:
Locomotion in leech takes place by:

  • Looping or Crawling movement: This type of movement is brought about, by the contraction and relaxation of muscles. The two suckers, serve for attachment, during movement on a substratum.
  • Swimming movement: Leeches swim very actively and perform undulating movements in the water.

Question 3.
Explain the male reproductive system of the rabbit with a labelled diagram.
Answer:
A pair of Testes, which are ovoid in shape, are enclosed by Scrotal Sac, in the abdominal cavity. Each testis consists of numerous fine tubules called Seminiferous tubules. This network of tubules leads into a coiled tubule called Epididymis, which lead into the sperm duct called vas deferens. The vas deferens join in the urethra, just below the Urinary Bladder. The Urethra runs backwards and passes into the penis.
Samacheer Kalvi 10th Science Solutions Chapter 13 Structural Organisation of Animals 1
The Prostate gland, Cowper’s gland and Perineal gland are the three accessory glands, whose secretions are involved in reproduction.

VIII. Higher Order Thinking Skills (HOTS) Questions

Question 1.
Arjun is studying in tenth standard. He was down with fever and went to meet the doctor. As he went to the clinic he saw a patient undergoing treatment for severe leech bite. Being curious, Arjun asked the doctor why leech bite was not felt as soon as it attaches to the skin ? What would have been the reply given by the doctor?
Answer:
The leech makes a wound with the jaws by making a rasping movement. The blood is sucked by the muscular pharynx. The salivary secretion is poured in the wound. They inject an anaesthetic substances that prevents the host from feeling their bite. The saliva contains an active substances called hirudin which prevents the coagulation of the blood.

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Question 2.
Shylesh has some pet animals at his home. He has few rabbits too, one day while feeding them he observed something different with the teeth. He asked his grandfather, why is it so? What would have been the explanation of his grandfather?
Answer:
The explanation of the grandfather would have been as follows: Teeth are hard bone – like structures, used to cut, tear and grind the food. There are incisors, canines, premolars and molars teeth are seen. Canines are absent. Something different from the teeth is the gap between incisors and premolar, which is called Diastema. It helps in mastication and chewing the food.

IX. Value-Based Questions.

Question 1.
Leeches do not have an secretion of digestive juices and enzymes -Why?
Answer:
The blood sucked by the leech is stored up in the crop. The blood gets haemolysed in the crop. Then the blood is passed drop by drop into the stomach where it is digested slowly by the peptolytic enzyme. The digested blood is absorbed slowly by the intestine.

Question 2.
How is the digestive system of rabbit suited for a herbivorous mode of feeding?
Answer:
The caecum is a thin – walled sac present at the Junction of the small intestine and large intestine. It contains bacteria, that helps in digestion of cellulose. So the digestive system of Rabbit is suited for a herbivorous mode of feeding.

Samacheer Kalvi 10th Science Structural Organisation of Animals Additional Questions Solved

I. Match the following.

Question 1.

1. Segmentation (a) Hirudin
2. Gregarious (b) Vagina
3. Blood clotting (c) Metamerism
4. Four chambered (d) Move-in groups
5. Oviducts (e) Septum

Answer:
1. (c) Metamerism
2. (d) Move-in groups
3. (a) Hirudin
4. (e) Septum
5. (b) Vagina.

II. Choose the correct pair.

Question 1.
The digestion of cellulose in rabbit takes place in:
(a) vemiform appendix
(b) colon
(c) caecum
(d) ileum
Answer:
(c) caecum

Question 2.
(a) Ureter and Dermis
(b) Clitella and Mouth
(c) Vagina and Egg case
(d) Sucker and Small intestine.
Answer:
(c) Vagina and Egg case

Question 3.
Total number of incisors teeth in rabbit is:
(a) 8
(b) 6
(c) 10
(d) 4
Answer:
(b) 6

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Question 4.
(a) Mouth and Metamerism
(b) Ganglion and Colon
(c) Intestine and Integument
(d) Expiration and Inspiration.
Answer:
(d) Expiration and Inspiration

Question 5.
(a) Sweat glands and Sebaceous glands
(b) Triradiate and Thoracic
(c) Trachea and Tricuspid
(d) Seminiferous and Sansuivorous.
Answer:
(a) Sweat glands and Sebaceous glands

III. Fill in the Blanks.

Question 1.
In leech, the ______ tissue lies beneath longitudinal muscles and fills the entire colon around the gut.
Answer:
Botryoidal.

Question 2.
_____ teeth are absent in Rabbit.
Answer:
Canine.

Question 3.
Digestion in leech takes place in the stomach by the action of ______ enzyme.
Answer:
Proteolytic.

Question 4.
The anterior part of the windpipe is enlarged to form ______ or _____.
Answer:
Larynx or Voicebox.

Question 5.
The opening of the pulmonary artery and Aorta are guarded by three ________ valves.
Answer:
Semilunar.

Question 6.
The coelomic fluid contains _____.
Answer:
Haemoglobin.

Question 7.
The Jaws of Leeches are provided with _____.
Answer:
Papillae.

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Question 8.
The pain of testes is enclosed by a sac called _____.
Answer:
Scrotal Sac.

Question 9.
Central Nervous System consists of _______ and _____.
Answer:
Brain, Spinal Chord.

Question 10.
Autonomous nervous system comprises _______ and _____.
Answer:
Sympathetic, Parasympathetic.

Question 11.
Each kidney is made up of several _____.
Answer:
Nephrons.

Question 12.
In Rabbits, as male and female sexes are separate, _____ is exhibited.
Answer:
Sexual dimorphism.

Question 13.
The other name for the voice box is _____ and the other name for the windpipe is _____.
Answer:
Larynx, Trachea.

Question 14.
The right and left ventricles are separated by _____.
Answer:
Inter Ventricular Septum

Question 15.
The external ear or ______ is situated at the top of the head in Rabbits.
Answer:
Pinnae.

IV. Choose the correct answer.

Question 1.
The largest portion of the Alimentary Canal in Leech is _____.
(a) Mouth
(b) Pharynx
(c) Oesophagus
(d) Crop.
Answer:
(d) Crop.

Question 2.
In Leech, the dense network of tiny blood vessels, containing Haemocoelic fluid is called _____.
(a) Artery
(b) Capillaries
(c) Vein
(d) Blood Vessel.
Answer:
(b) Capillaries

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Question 3.
Body wall of Leech includes ______ layers.
(a) 2
(b) 3
(c) 5
(d) 4.
Answer:
(c) 5

Question 4.
In female Rabbits, these four or five structures are present on the ventral surface between the thorax and abdomen _____.
(a) Nipples or Trets
(b) Limbs
(c) Vibrissae
(d) Diaphragm.
Answer:
(a) Nipples or Trets

Question 5.
In male rabbits, this structure is found in the ventral side of Anus _____.
(a) Claws
(b) Sweat glands
(c) Penis
(d) Abdomen.
Answer:
(c) Penis

V. Read the following statements and correct them, if it is not true.

Question 1.
Dorsal surface of Leech is Orange Yellow or Red in colour.
Answer:
False.
Correct Statement: Dorsal surface of Leech is Olive green in colour.

Question 2.
The clitellum is formed on segments 9 to 11, which is meant to produce cocoon, during the breeding season.
Answer:
True.

Question 3.
The lower side of the thoracic cavity in Rabbit is the dome-shaped Sternum.
Answer:
False.
Correct Statement: The lower side of the thoracic cavity in Rabbit is the dome-shaped Diaphragm.

Question 4.
Mammary glands are the modified glands of the skin.
Answer:
True.

Question 5.
The supra pharyngeal ganglion lies below the Pharynx and is formed by the fusion of four pairs of Ganglia.
Answer:
False.
Correct Statement: The sub pharyngeal ganglion lies below the Pharynx and is formed by the fusion of four pairs of Ganglia.

VI. Find the Odd One Out.

Question 1.
Cuticle, Epidermis, Sphincters, Muscular layer.
Answer:
Sphincters.

Question 2.
Undigested food, Genital pore, Rectum, Anus.
Answer:
Genital pore.

Question 3.
Gregarious, Thorax, Abdomen, Pinnae.
Answer:
Gregarious.

Question 4.
Nephrons, Nitrogenous waste, Scrotal Sac, Urea.
Answer:
Scrotal Sac.

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Question 5.
Teats or Nipples, Hairs, Claws, Nails
Answer:
Teats or Nipples.

V. Answer the following in a word or sentence.

Question 1.
Where is mouth located in Leech?
Answer:
The mouth is located in the middle of the Anterior Sucker.

Question 2.
List out the structures derived from the skin of rabbit.
Answer:
The structures derived from the skin of rabbit are hair, claws, nails and glands like sweat glands, sebaceous glands and mammary glands.

Question 3.
What is Epiglottis?
Answer:
Epiglottis is a flap, in the neck, which prevents the entry of food into the trachea through the glottis.

Question 4.
The excretory system of rabbit is called a urinogenital system?
Answer:
The Urinogenital system of rabbit comprises the urinary system and genital or reproductive system. So they are called urinogenital system.

Question 5.
Where are Annular and Segmental receptors located?
Answer:
Annular receptors are located in each annulus and Segmental receptors are located on the first annulus of each segment.

Question 6.
What are Vibrissae?
Answer:
In Rabbit, from each side of upper lip tactile hairs or Whiskers present, which are called Vibrissae.

Question 7.
What is the scientific name of rabbit?
Answer:
The scientific name of rabbit is Oryctolagus cuniculus.

Question 8.
Which glands regulate the body temperature of Rabbits?
Answer:
The Sweat glands and Sebaceous glands embedded in the skin regulate the body temperature of Rabbits.

VI. Answer Briefly.

Question 1.
Explain the Excretory System of Leech briefly.
Answer:
In Leech, excretion takes place by segmentally arranged paired tubules called Nephridia. There are 17 pairs of Nephridia, which open out by nephridiopores from 6th to 22nd segments.

Question 2.
Explain the types of coelom in leech.
Answer:
Leech contains special longitudinal canals called haemocoelic canals filled with a blood like fluid called haemocolic fluid. This type of coelom is called Haemocoel.

Question 3.
What are Receptors? Write a short note on Receptors.
Answer:
Sensory projections are called Receptors on the dorsal side, there are five pairs of eyes on the first five segments. Annular receptors are located in each Annulus and segmental receptors are located on the first annulus of each segment.

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Question 4.
What is hermaphrodite?
Answer:
If the male and female reproductive organs are present in the same animals, ‘ then the organism is said to hermaphrodite.

Question 5.
Name three accessory glands, which involve in the male reproductive system of Rabbit.
Answer:

  • Prostate gland
  • Cowper’s gland
  • Perineal gland.

Their secretion involved in Reproduction.

Question 6.
Name the following:
(a) Three membranes, which cover the Brain in Rabbit
(b) Division of brain
Answer:
(a) Three membranes, which cover the Brain in Rabbit

  • Dura mater
  • Inner Piameter
  • Arachnoid membrane.

(b) Division of brain

  1. Forebrain (Prosencephalon)
  2. Midbrain (Mesencephalon)
  3. Hindbrain (Rhombencephalon).

Question 7.
Write a note on Medicinal value of Leech.
Answer:

  1. Leeches breaks up blood clots.
  2. Increases blood circulation
  3. Leeches are used to treat cardio vascular diseases.
  4. Saliva of leeches are used in the preparation of drugs that can treat hypertension.

Question 8.
With the Tabular Column, mention the Divisions of the body of Leech.
Answer:

Region Segments
Cephalic region 1st – 5th
Pre-clitellar region 6th, 7th and 8th
Clitellar region 9th, 10th and 11th
Middle region 12th – 22nd
Caudal region 23rd – 26th
Posterior sucker 27th – 33rd

Question 9.
Explain the hermaphrodite structure of leech.
Answer:
Leech is hermaphrodite because it contain both male and female genital organ in same animal.
Male Reproductive system: It is formed of: (i) Testes, (ii) Vas efferens, (Hi) Epididymis, (iv) Ejaculatory ducts and atrium.

There are eleven pairs of testes, one pair in each segment from 12 to 22 segments. They are in the form of spherical sacs called testes sacs. From each testes arises a short duct called vas efferens, which join with the vas deferens. The vas deferens becomes convoluted to form the epididymis or sperm vesicle, to store spermatozoa.

The epididymis leads to a short duct called ejaculatory duct. The ejaculatory ducts on both sides join to form the genital atrium. The atrium consists of two regions, the coiled prostate glands and the penial sac consisting of penis that opens through the male genital pore.

Female Reproductive system : It consists of ovaries, oviducts and vagina. There is a single pair of ovary in the 11th segment on the ventral side. Each ovary is a coiled ribbon-shaped structure. The ova are budded off from the ovary. From each ovary runs a short oviduct. The oviducts of the two sides joins together, to form a common oviduct. The common oviduct opens into a pear-shaped vagina which lies mid-ventrally in the posterior part of the 11th segment.

VIII. Label the following diagrams.

Question 1.
Draw and label the female Reproductive System of Rabbit.
Answer:
Samacheer Kalvi 10th Science Solutions Chapter 13 Structural Organisation of Animals 2

Question 2.
Draw and label the Digestive System of Leech.
Answer:
Samacheer Kalvi 10th Science Solutions Chapter 13 Structural Organisation of Animals 3

Question 3.
Draw and label the Brain of the rabbit.
Answer:
Samacheer Kalvi 10th Science Solutions Chapter 13 Structural Organisation of Animals 4

Question 4.
Draw and label the Lung of Rabbit.
Answer:
Samacheer Kalvi 10th Science Solutions Chapter 13 Structural Organisation of Animals 5

IX. Answer the following in Detail.

Question 1.
Explain the Digestive System of Rabbit with a neat labelled diagram.
Answer:
Teeth are hard, bone – like structures used to cut, tear and grind the food materials. The digestive system includes the Alimentary canal and the associated digestive glands. The alimentary canal consists of a mouth, buccal cavity, pharynx, oesophagus, stomach, small intestine and anus. The mouth is a transverse slit, bounded by upper and lower lips.

It leads to a buccal cavity muscular tongue is at the floor of the buccal cavity. Jaws bear teeth. Buccal cavity leads into Oesophagus. Oesophagus opens into the stomach followed by the small intestine. The thin-walled Sac, called Caecum, present at the Junction of the small intestine and large intestine, contains bacteria, which helps in digestion of cellulose.
Samacheer Kalvi 10th Science Solutions Chapter 13 Structural Organisation of Animals 6
The digestive glands such as salivary glands, gastric glands, Liver, Pancreas and intestinal glands secrete digestive juices, which help in the digestion of food in the Alimentary canal. After digestion, from the small intestine, the undigested food and wastes enter into the large intestine, which has colon and rectum. The rectum opens outside by the Anus.

Question 2.
Explain the male and the female Reproductive system of Leech with a neat labelled diagram.
Answer:
Leech is hermaphrodite, both the male and female reproductive organs are present in the same.
Male Reproductive system: There are eleven pairs of testes, one pair in each segment from 12 to 22 segments. They are in the form of Spherical Sacs called testes sacs. From each testis arises a short duct called Vas efferents, which join with the Vas deferens.

The Vas deferens becomes convoluted to form the epididymis or sperm vesicle, to store spermatozoa. The epididymis leads to a short duct called the ejaculatory duct. The ejaculatory duct on both sides joins to form Genital Atrium. The Atrium consists of prostate glands and the Penial sac, with Penis, that opens through the genital pore.

Female Reproductive system: It consists of Ovaries, Oviducts and Vagina. Each ovary is coiled, ribbon-shaped, and the single pair of the ovary is in the 11th segment. The Ova are budded off from the ovary. From each ovary runs a short oviduct, and the oviducts of two sides join together, to form a common Oviduct, which opens into a pear-shaped Vagina. Vagina lies mid ventrally in the posterior part of 11th segment.

Internal fertilisation takes place. This is followed by cocoon formation. The cocoon or egg case is formed around 9th, 10th and 11th segments. Development directs, proceeds in a cocoon, which contain one to 24 embryos. The emerging young leech resembles the adult Leech.
Samacheer Kalvi 10th Science Solutions Chapter 13 Structural Organisation of Animals 7

Question 3.
Explain the Circulatory system of Rabbit with a neat labelled diagram of a Heart.
(i) Observe the external morphology of leech specimen in your biology laboratory.
(ii) Can you find leeches in your locality?
(iii) In which geographical areas are leeches found more predominantly in India?
Answer:
The Circulatory system consists of blood, blood vessels and Heart. The Heart is pear-shaped, lies in between the lungs, in the thoracic cavity. The heart is covered by a double-layered Pericardium. The Heart is four-chambered, with two Auricles and two ventricles. The right and the left ventricles are separated by inter auricular septum. The right and the left ventricles are separated by the interventricular septum. The right auricle opens into the right ventricle, by right Auriculoventricular aperture, guarded by a tricuspid valve. The left auricle opens into the left ventricle by left Auriculoventricular aperture, guarded by a bicuspid valve or Mitral valve. The opening of the Pulmonary Artery and Aorta are guarded by three semilunar valves.

The right Auricle receives deoxygenated blood through two Precaval (superior vena cava) and one postcaval (Inferior vena cava) veins from all parts of the body. The left auricle receives oxygenated blood from the pulmonary veins from the lungs.
From the right ventricle arises Pulmonary trunk, which carries the deoxygenated blood to the lungs and from the left ventricle arises the systemic arch (aorta) which supplies Oxygenated blood to all parts of the body.
Samacheer Kalvi 10th Science Solutions Chapter 13 Structural Organisation of Animals 8
(i) Observe the external morphology of Leech in the preserved bottle specimens in the Biology Laboratory.
(ii) No, Not in our areas. The Leeches are common in hilly areas
(iii) Most Leech species are found in shallow, slow-moving freshwater and moist soil on land.

IX. Higher Order Thinking Skills (HOTS) Questions

Question 1.
How were Leeches used in Ancient times?
Answer:
Leeches were used in medicines from Ancient times. Until the 19th century, Leeches were used to draw blood from patients.

Question 2.
How are Leeches used in modern times?
Answer:
In modem times, Leeches find medical use in the treatment of Joint diseases such as Epicondylitis and Osteoarthritis, Extremity vein diseases and Microsurgery. Hirudin is a valuable dmg for some blood clotting disorders. There are doctors who will use Leeches to treat muscle cramps.

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Question 3.
What is the benefit of Rabbit?
Answer:
Rabbit meat is white meat of high quality, easily digestible with low fat, low cholesterol and high protein compared to most other meats. It is an excellent source of vitamins B3 and B12, minerals, and trace elements (Phosphorus, Potassium and Selenium) Rabbit meat is an excellent balance of fatty acids, Omega 3 than chicken or pork.

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