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NCERT Solutions

Atoms and Molecules

Madhya Pradesh Board · Class 9 · Science

NCERT Solutions for Atoms and Molecules — Madhya Pradesh Board Class 9 Science.

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An infographic summarizing the key postulates of Dalton's Atomic Theory, which laid the foundation for modern chemistry.
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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:
Percentage by weight=Mass of elementTotal mass of compound×100\text{Percentage by weight} = \frac{\text{Mass of element}}{\text{Total mass of compound}} \times 100

Step 1: Percentage of Boron
% Boron=0.0960.24×100=40%\% \text{ Boron} = \frac{0.096}{0.24} \times 100 = 40\%

Step 2: Percentage of Oxygen
% Oxygen=0.1440.24×100=60%\% \text{ Oxygen} = \frac{0.144}{0.24} \times 100 = 60\%

Verification: 40%+60%=100%40\% + 60\% = 100\%

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

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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:
C : O=3.0:8.0=3:8\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= 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 g= 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.

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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 | NH4+\text{NH}_4^+ | +1+1 |
| Hydroxide ion | OH\text{OH}^- | 1-1 |
| Carbonate ion | CO32\text{CO}_3^{2-} | 2-2 |
| Sulphate ion | SO42\text{SO}_4^{2-} | 2-2 |
| Nitrate ion | NO3\text{NO}_3^- | 1-1 |
| Phosphate ion | PO43\text{PO}_4^{3-} | 3-3 |

These ions participate in ionic bonding just like monoatomic ions.

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4Write the chemical formulae of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate
5Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate
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

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Frequently Asked Questions

What are the important topics in Atoms and Molecules for Madhya Pradesh Board Class 9 Science?
Atoms and Molecules covers several key topics that are frequently asked in Madhya Pradesh Board Class 9 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.
How to score full marks in Atoms and Molecules — Madhya Pradesh Board Class 9 Science?
Understand the core concepts first, then work through the 29 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.
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