Chapter 3 of 12

Updated 6 August 2026

NCERT Solutions

Atoms and Molecules — NCERT Solutions

Madhya Pradesh Board · Class 9 · Science

NCERT Solutions for Atoms and Molecules, Madhya Pradesh Board Class 9 Science: 6 textbook questions solved step by step. Covers Exercises.

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An infographic summarizing the key postulates of Dalton's Atomic Theory, which laid the foundation for modern chemistry. — An illustration from Super Tutor's Atoms and Molecules chapter — alongside flashcards, concept maps and practice questions.

6 Questions Solved · 1 Section

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Exercises

1A 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.Show solution

Given:

Total mass of compound = 0.24 g
Mass of boron = 0.096 g
Mass of oxygen = 0.144 g

Formula used:
$\text{Percentage by weight} = \frac{\text{Mass of element}}{\text{Total mass of compound}} \times 100$

Step 1: Percentage of Boron
$\% \text{ Boron} = \frac{0.096}{0.24} \times 100 = 40\%$

Step 2: Percentage of Oxygen
$\% \text{ Oxygen} = \frac{0.144}{0.24} \times 100 = 60\%$

Verification: $40\% + 60\% = 100\%$ ✓

Answer: The compound contains 40% Boron and 60% Oxygen by weight.

2When 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?Show solution

Given (first case):

Mass of carbon = 3.0 g
Mass of oxygen = 8.00 g
Mass of CO₂ produced = 11.00 g

Observation from first case:
Carbon and oxygen combine in the ratio:
$\text{C : O} = 3.0 : 8.0 = 3 : 8$
This means 3 g of carbon requires exactly 8 g of oxygen to produce 11 g of CO₂.

Second case:

Mass of carbon = 3.00 g
Mass of oxygen available = 50.00 g

Step 1: Find oxygen required to burn 3.00 g of carbon completely.
From the fixed ratio, oxygen required $= 8.00$ g.

Step 2: Since 50.00 g of oxygen is available but only 8.00 g is needed, oxygen is in excess. Carbon is the limiting reactant.

Step 3: Mass of CO₂ formed
$= 3.00 + 8.00 = 11.00 \text{ g}$

Answer: 11.00 g of carbon dioxide will be formed.

Law governing this: This is governed by the Law of Constant Proportions (Law of Definite Proportions), which states that in a chemical compound, elements are always present in definite proportions by mass, regardless of the source or method of preparation.

3What are polyatomic ions? Give examples.Show solution

Definition:
A polyatomic ion is a group of atoms carrying a net electric charge (positive or negative). These atoms are covalently bonded together and the group as a whole acts as a single ion.

Examples:

Ion	Formula	Charge
Ammonium ion	$\text{NH}_4^+$	$+1$
Hydroxide ion	$\text{OH}^-$	$-1$
Carbonate ion	$\text{CO}_3^{2-}$	$-2$
Sulphate ion	$\text{SO}_4^{2-}$	$-2$
Nitrate ion	$\text{NO}_3^-$	$-1$
Phosphate ion	$\text{PO}_4^{3-}$	$-3$

These ions participate in ionic bonding just like monoatomic ions.

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

Concept: Chemical formulae are written by criss-crossing the valencies of the combining ions.

(a) Magnesium chloride

Magnesium ion: $\text{Mg}^{2+}$ (valency = 2)
Chloride ion: $\text{Cl}^-$ (valency = 1)
Criss-cross: Formula = $\mathbf{MgCl_2}$

(b) Calcium oxide

Calcium ion: $\text{Ca}^{2+}$ (valency = 2)
Oxide ion: $\text{O}^{2-}$ (valency = 2)
Criss-cross and simplify (ratio 2:2 → 1:1): Formula = $\mathbf{CaO}$

(c) Copper nitrate

Copper ion: $\text{Cu}^{2+}$ (valency = 2)
Nitrate ion: $\text{NO}_3^-$ (valency = 1)
Criss-cross: Formula = $\mathbf{Cu(NO_3)_2}$

(d) Aluminium chloride

Aluminium ion: $\text{Al}^{3+}$ (valency = 3)
Chloride ion: $\text{Cl}^-$ (valency = 1)
Criss-cross: Formula = $\mathbf{AlCl_3}$

(e) Calcium carbonate

Calcium ion: $\text{Ca}^{2+}$ (valency = 2)
Carbonate ion: $\text{CO}_3^{2-}$ (valency = 2)
Criss-cross and simplify (ratio 2:2 → 1:1): Formula = $\mathbf{CaCO_3}$

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5Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate

(a) Quick lime

Chemical formula: $\text{CaO}$
Elements present: Calcium (Ca) and Oxygen (O)

(b) Hydrogen bromide

Chemical formula: $\text{HBr}$
Elements present: Hydrogen (H) and Bromine (Br)

(c) Baking powder

Baking powder contains Sodium hydrogen carbonate ( $\text{NaHCO}_3$ ) as the main active ingredient.
Elements present: Sodium (Na), Hydrogen (H), Carbon (C), and Oxygen (O)

(d) Potassium sulphate

Chemical formula: $\text{K}_2\text{SO}_4$
Elements present: Potassium (K), Sulphur (S), and Oxygen (O)

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6Calculate 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

Concept: Molar mass = Sum of atomic masses of all atoms in the molecular formula.

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

(a) Ethyne, $\text{C}_2\text{H}_2$
$M = (2 \times 12) + (2 \times 1)$
$M = 24 + 2 = \mathbf{26 \text{ g mol}^{-1}}$

(b) Sulphur molecule, $\text{S}_8$
$M = 8 \times 32$
$M = \mathbf{256 \text{ g mol}^{-1}}$

(c) Phosphorus molecule, $\text{P}_4$
$M = 4 \times 31$
$M = \mathbf{124 \text{ g mol}^{-1}}$

(d) Hydrochloric acid, $\text{HCl}$
$M = 1 + 35.5$
$M = \mathbf{36.5 \text{ g mol}^{-1}}$

(e) Nitric acid, $\text{HNO}_3$
$M = 1 + 14 + (3 \times 16)$
$M = 1 + 14 + 48 = \mathbf{63 \text{ g mol}^{-1}}$

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Frequently Asked Questions

What are the important topics in Atoms and Molecules for Madhya Pradesh Board Class 9 Science?

Key topics in Atoms and Molecules include Historical Background and Laws of Chemical Combination, Dalton's Atomic Theory, Atoms - Structure and Properties, Molecules and Their Types. Study these first, then practise questions on each for Class 9 exams.

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How should I revise Atoms and Molecules for Class 9 exams?

Learn the core ideas first, then work through the 29 practice questions on Atoms and Molecules. Revise definitions regularly and use flashcards for quick recall before the exam.

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