NCERT Solutions for Class 9 Science Chapter 3 Atoms and Molecules

1. A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.

Answer: Given – Mass of boron = 0.096 g, Mass of oxygen = 0.144g, Mass of sample = 0.24g.

Percentage of boron by weight in a compound = ( 0.096 / 0.24 ) x 100 % = 40%

Percentage of oxygen by weight in a compound = ( 0.144 / 0.24 ) x 100 % = 60%

2. When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?

Answer: Given – 3 g of carbon reacts with 8 g of oxygen to form 11 g of carbon dioxide.

If 3 g of carbon is burnt in 50 g of oxygen, thus 8 gram of oxygen will react with 3 gram of carbon, therefore, 42 grams of oxygen will be left unreactive. Also, 11 g of carbon dioxide will be formed in this case. This answer is governed by law of constant proportion.

3. What are polyatomic ions? Give examples.

Answer: Group of atoms having charge on them is called polyatomic ions. Examples of polyatomic ion are ammonium ion NH₄⁺, hydroxide ion OH^–, carbonate ion CO₃^2-, nitrate ion NO₃^–.

4. Write the chemical formulae of the following – (a) Magnesium chloride (b) Calcium oxide (c) Copper nitrate (d) Aluminium chloride (e) Calcium carbonate

Answer:

• Magnesium chloride – MgCl₂
• Calcium oxide – CaO
• Copper nitrate – Cu(NO₃)₂
• Aluminium chloride – AlCl₃
• Calcium carbonate – CaCO₃

5. Give the names of the elements present in the following compounds –

(a) Quick lime (b) Hydrogen bromide (c) Baking powder (d) Potassium sulphate

Answer:

• Quick lime – CaO – Calcium, Oxygen
• Hydrogen bromide – HBr – Hydrogen, Bromine
• Baking powder – NaHCO₃ – Sodium, Hydrogen, Carbon, Oxygen
• Potassium sulphate – K₂SO₄ – Potassium, Sulphur, Oxygen

6. Calculate the molar mass of the following substances- (a) Ethyne, C₂H₂ (b) Sulphur molecule, S₈ (c) Phosphorus molecule, P₄ (Atomic mass of phosphorus = 31)(d) Hydrochloric acid, HCl (e) Nitric acid, HNO₃

Answer: (a) Ethyne, C₂H₂: Molar mass of ethyne C₂H₂ = 2 x 12 + 2 x 1 = 28 g

(b) Sulphur molecule, S₈: Molar mass of sulphur molecule S₈ = 8 x 32 = 256 g

(c) Phosphorus molecule, P₄:Molar mass of phosphorus molecule P₄ = 4 x 31 = 124 g

(d) Hydrochloric acid, HCl: Molar mass of hydrochloric acid HCl = 1 + 35.5 = 36.5 g

(e) Nitric acid, HNO₃: Molar mass of nitric acid HNO₃ = 1 + 14 + 3 x 16 = 63 g

7. What is the mass of—

(a) 1 mole of nitrogen atoms?

Answer: Mass of 1 mole of nitrogen is 14 g

(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?

Answer: Mass of 4 moles of aluminium atoms = 4 x 27 = 108 g

(c) 10 moles of sodium sulphite (Na2SO3)?

Answer: Mass of 10 moles of sodium sulphite = 10 x ( 2 x 23 + 32 + 3 x 16) g = 10 x 126 g = 1260 g

8. Convert into mole.

(a) 12 g of oxygen gas

Answer: 12 g of oxygen gas –
32 g of oxygen gas = 1 mole
∴ 12 grams of oxygen gas = 12 / 32 = 0.375 moles

(b) 20 g of water

Answer: 20 g of water –
18 g of water = 1 mole
∴ 20 g of water = 20 / 18 = 1.1 moles

(c) 22 g of carbon dioxide

Answer: 22 g of carbon dioxide
44 g of carbon dioxide = 1 mole
∴ 22 g of carbon dioxide = 22 /44 = 0.5 moles

9. What is the mass of:

(a) 0.2 mole of oxygen atoms?

Answer: 0.2 mole of oxygen atoms –
Mass of one mole of oxygen atoms is 16 g
∴ Mass of 0.2 mole of oxygen atoms = 0.2 x 16 g = 3.2 g

(b) 0.5 mole of water molecules?

Answer: 0.5 mole of water molecules –
Mass of one mole of water molecules is 18 g
∴ mass of 0.5 mole of water molecules = 0.5 x 18 g = 9 g

10. Calculate the number of molecules of sulphur (S₈) present in 16 g of solid sulphur.

Answer: 1 mole of sulphur S₈ = 8 x 32 = 256 g
∴ 256 g of solid sulphur contains = 6.022 x 10²³ molecules
∴ 16 g of solid sulphur will have = ( 6.022 x 10²³ / 256 ) x 16 molecules
≅3.76 x 10²² molecules

11. Calculate the number of aluminium ions present in 0.051 g of aluminium oxide. (Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u)

Answer: 1 mole of aluminium oxide Al₂O₃ = 2 x 27 + 3 x 16 = 102 g
∴ 102 g of Al₂O₃ = 6.022 x 10²³ Al₂O₃ molecules.
Then, 0.051 g of aluminium oxide contains = ( 6.022 x 10²³ / 102 ) x 0.051 = 3.011 x 10²⁰ molecules.
2 aluminium ions are present in one molecule of aluminium oxide
∴ number of aluminium ion present 3.011 x 10²⁰ molecules for 0.051 g of aluminium oxide = 2 x 3.011 x 10²⁰ = 6.022 x 10²⁰

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