Case Study Questions for Class 9 Science Chapter 9 Atomic Foundations of Matter (Exploration Book) 2026-27

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This page provides Case Study Questions for Class 9 Science Chapter 9 – Atomic Foundations of Matter from the latest NCERT Exploration textbook. Each set contains a reading passage followed by 4 objective/short-answer questions, exactly as expected in the CBSE Board examination pattern.

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Exam Tip: Case Study questions carry 4 marks each in the CBSE Class 9 Science paper. Read the passage carefully before attempting the MCQs or short answers.

1

Case Study: Laws of Chemical Combination

Read the passage carefully, then answer all four questions
Ch 9 · Atomic Foundations of Matter

In the study of chemical reactions, Antoine L. Lavoisier established the Law of Conservation of Mass, which states that mass can neither be created nor destroyed in a chemical reaction. The total mass of the reactants is always equal to the total mass of the products. Following this, Joseph Proust formulated the Law of Definite Proportions (or Constant Proportions), which states that in a chemical substance, the elements are always present in definite proportions by mass.

For example, water (H2O) always contains hydrogen and oxygen in the ratio of 1:8 by mass, regardless of the source of the water. Thus, if 9 g of water is decomposed, 1 g of hydrogen and 8 g of oxygen are always obtained. These foundational laws paved the way for Dalton’s atomic theory, which provided a theoretical explanation for these experimental observations.

1
What is the ratio by mass of nitrogen to hydrogen in ammonia (NH3)? (Given atomic mass of N = 14 u, H = 1 u)
a 1:3
b 14:3
c 3:14
d 1:14
Correct Answer (b) 14:3 ✅
Explanation

An ammonia molecule (NH3) consists of one nitrogen atom and three hydrogen atoms. The mass of one nitrogen atom is 14 u, and the mass of three hydrogen atoms is 3 × 1 = 3 u. Therefore, the elements nitrogen and hydrogen are always present in the ratio of 14:3 by mass.

2
If 100 g of calcium carbonate (CaCO3) decomposes completely to produce 56 g of calcium oxide (CaO), what mass of carbon dioxide (CO2) is produced?
a 56 g
b 100 g
c 44 g
d 156 g
Correct Answer (c) 44 g ✅
Explanation

According to the Law of Conservation of Mass, the total mass of the reactants must equal the total mass of the products. Mass of reactant (100 g) = Mass of CaO (56 g) + Mass of CO2. Therefore, the mass of CO2 is 100 g – 56 g = 44 g.

3
Which scientist formulated the Law of Definite Proportions?
a Antoine Lavoisier
b John Dalton
c Joseph Proust
d J.J. Thomson
Correct Answer (c) Joseph Proust ✅
Explanation

The Law of Definite Proportions, which states that elements in a chemical compound are always present in a fixed ratio by mass, was formulated by the French chemist Joseph Proust.

4
According to Dalton’s Atomic Theory, which postulate directly explains the Law of Conservation of Mass?
a Atoms of a given element are identical in mass and chemical properties.
b Atoms combine in the ratio of small whole numbers to form compounds.
c Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction.
d All matter is made of very tiny particles called atoms.
Correct Answer (c) Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction. ✅
Explanation

The postulate stating that atoms are indivisible particles that cannot be created or destroyed is the theoretical justification for the Law of Conservation of Mass. Since atoms merely rearrange during a reaction, the total mass remains constant.

2

Case Study: Ions and Chemical Formulae

Read the passage carefully, then answer all four questions
Ch 9 · Atomic Foundations of Matter

Compounds composed of metals and non-metals contain charged species known as ions. A negatively charged ion is called an anion, and a positively charged ion is called a cation. Some ions consist of a single charged atom, while others consist of a group of atoms carrying a net charge, known as a polyatomic ion.

The chemical formula of a compound is a symbolic representation of its composition. To write the chemical formula of a compound, we first write the symbols of the constituent elements and their valencies. Then, we cross-over the valencies of the combining atoms. For ionic compounds, the symbol of the metal is written first, followed by the non-metal. If the compound contains a polyatomic ion and requires more than one of that ion to balance the charges, the polyatomic ion is enclosed in brackets before writing the subscript number.

1
What is the correct chemical formula for magnesium chloride?
a MgCl
b Mg2Cl
c MgCl2
d Mg2Cl2
Correct Answer (c) MgCl2 ✅
Explanation

Magnesium (Mg) has a valency of +2, and chlorine (Cl) has a valency of -1. When we cross-over these valencies, one atom of magnesium combines with two atoms of chlorine, resulting in the formula MgCl2.

2
In the chemical formula of aluminium sulfate, what is the number of sulfate ions required to balance the charge of the aluminium ions?
a 1
b 2
c 3
d 4
Correct Answer (c) 3 ✅
Explanation

Aluminium (Al) has a valency of +3, and the sulfate polyatomic ion (SO4) has a valency of -2. Crossing the valencies over yields the formula Al2(SO4)3, meaning three sulfate ions are required for every two aluminium ions.

3
Which of the following represents a polyatomic ion?
a Chloride (Cl–)
b Oxide (O2-)
c Ammonium (NH4+)
d Sodium (Na+)
Correct Answer (c) Ammonium (NH4+) ✅
Explanation

A polyatomic ion is a group of atoms carrying a net charge. Ammonium (NH4+) is composed of one nitrogen atom and four hydrogen atoms carrying a net positive charge, making it a polyatomic ion.

4
What is the simplified chemical formula for calcium oxide?
a Ca2O2
b CaO
c CaO2
d Ca2O
Correct Answer (b) CaO ✅
Explanation

Calcium (Ca) has a valency of +2 and oxygen (O) has a valency of -2. When the valencies are identical, they simplify to a 1:1 ratio. Therefore, the formula is written simply as CaO.

Chapters covered in CBSE Class 9 Science Latest Book – Exploration

  • Chapter 1: Exploration: Entering the World of Secondary Science
  • Chapter 2: Cell: The Building Block of Life
  • Chapter 3: Tissues in Action
  • Chapter 4: Describing Motion Around Us
  • Chapter 5: Exploring Mixtures and their Separation
  • Chapter 6: How Forces Affect Motion
  • Chapter 7: Work, Energy, and Simple Machines
  • Chapter 8: Journey Inside the Atom
  • Chapter 9: Atomic Foundations of Matter
  • Chapter 10: Sound Waves: Characteristics and Applications
  • Chapter 11: Reproduction: How Life Continues
  • Chapter 12: Patterns in Life: Diversity and Classification
  • Chapter 13: Earth as a System: Energy, Matter, and Life

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