Class 9th Science Chapter-3 Exercises

Atoms and Molecules

Exercises – Answers and Solutions

---

Question 1

A 0.24 g sample of a compound of oxygen and boron contains 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition by weight.

Solution:

Total mass of compound = 0.24 g

Percentage of boron:

$$\frac{0.096}{0.24}\times 100=40\mathrm{\%}$$

Percentage of oxygen:

$$\frac{0.144}{0.24}\times 100=60\mathrm{\%}$$

Answer:

• Boron = 40%

• Oxygen = 60%

---

Question 2

When 3.0 g of carbon is burnt in 8.0 g of oxygen, 11.0 g of carbon dioxide is formed.
What mass of carbon dioxide will be formed when 3.0 g of carbon is burnt in 50.0 g of oxygen?
Which law governs this?

Solution:

Carbon reacts with oxygen in a fixed ratio.

Given:

• 3 g carbon + 8 g oxygen → 11 g CO₂

Even though oxygen is present in excess (50 g), only 8 g oxygen will react with 3 g carbon.

So,

$$\text{Mass of CO₂ formed}=11\text{ g}$$

Law governing this:

Law of Constant (Definite) Proportions

Answer:

• Mass of CO₂ formed = 11 g

• Law: Law of Constant Proportions

---

Question 3

What are polyatomic ions? Give examples.

Answer:

Polyatomic ions are groups of two or more atoms that carry a net charge and act as a single ion.

Examples:

• Ammonium (NH₄⁺)

• Hydroxide (OH⁻)

• Carbonate (CO₃²⁻)

• Sulphate (SO₄²⁻)

---

Question 4

Write the chemical formulae of the following:

---

Question 5

Give the names of the elements present in the following compounds:

(a) Quick lime (CaO)

• Calcium

• Oxygen

(b) Hydrogen bromide (HBr)

• Hydrogen

• Bromine

(c) Baking powder

• Sodium

• Hydrogen

• Carbon

• Oxygen

(d) Potassium sulphate (K₂SO₄)

• Potassium

• Sulphur

• Oxygen

---

Question 6

Question:
Calculate the molar mass of the following substances.

(Atomic masses used: H = 1, C = 12, N = 14, O = 16, S = 32, P = 31, Cl = 35.5)

---

(a) Ethyne (C₂H₂)

$$(2\times 12)+(2\times 1)=24+2=\boxed{{\displaystyle 26{\text{g mol}}^{-1}}}$$

---

(b) Sulphur molecule (S₈)

$$8\times 32=\boxed{{\displaystyle 256{\text{g mol}}^{-1}}}$$

---

(c) Phosphorus molecule (P₄)

$$4\times 31=\boxed{{\displaystyle 124{\text{g mol}}^{-1}}}$$

---

(d) Hydrochloric acid (HCl)

$$1+35.5=\boxed{{\displaystyle 36.5{\text{g mol}}^{-1}}}$$

---

(e) Nitric acid (HNO₃)

$$1+14+(3\times 16)=1+14+48=\boxed{{\displaystyle 63{\text{g mol}}^{-1}}}$$