Chapter: 9 – Atomic Foundations of Matter
Maximum Marks: 50
Suggested Time: 1 Hour 30 Minutes

This Class 9 Science Exploration Chapter 9 Sample Paper 1 is designed to help students practise important concepts from “Atomic Foundations of Matter.” The paper covers laws of chemical combination, atoms and molecules, atomic mass, molecular mass, chemical formulae, valency, ions, and basic calculations involving atoms and molecules. Complete solutions are provided at the end.

Sample Paper 1

General Instructions

1. All questions are compulsory.
2. Read each question carefully.
3. Write chemical symbols and formulae correctly.
4. Show all steps in numerical questions.
5. Use appropriate scientific terminology.
6. Use the atomic masses given in the question where required.

Section A — Objective Questions

10 × 1 = 10 Marks

Q1. The law of conservation of mass states that mass:

a) Can be created during a chemical reaction
b) Can be destroyed during a chemical reaction
c) Is neither created nor destroyed in a chemical reaction
d) Always increases during a reaction

Q2. The smallest particle of an element that retains the chemical identity of the element is called:

a) Molecule
b) Atom
c) Ion
d) Compound

Q3. Which of the following represents a molecule of oxygen?

a) O
b) O₂
c) O₃
d) 2O

Q4. The formula of water is:

a) H₂O
b) H₂O₂
c) HO₂
d) H₃O

Q5. The combining capacity of an atom is called its:

a) Atomic mass
b) Valency
c) Density
d) Atomic number

Q6. Which of the following is a compound?

a) Oxygen
b) Iron
c) Water
d) Nitrogen

Q7. The formula of sodium chloride is:

a) NaCl
b) Na₂Cl
c) NaCl₂
d) Na₂Cl₂

Q8. A positively charged ion is called a:

a) Anion
b) Cation
c) Molecule
d) Neutron

Q9. The relative molecular mass of a substance is obtained by:

a) Adding the atomic masses of all atoms in its molecule
b) Multiplying atomic number by valency
c) Subtracting atomic masses
d) Dividing atomic mass by valency

Q10. According to the law of definite proportions, a pure compound always contains:

a) The same elements in the same fixed proportion by mass
b) Different elements in any proportion
c) Only one element
d) Equal numbers of atoms of all elements

Section B — Very Short Answer Questions

5 × 2 = 10 Marks

Q11. State the law of conservation of mass.

Q12. What is an atom? What is a molecule?

Q13. Define valency. What is the valency of oxygen in water?

Q14. Differentiate between a cation and an anion.

Q15. State the law of definite proportions with a suitable example.

Section C — Short Answer and Formula-Based Questions

4 × 3 = 12 Marks

Q16. Explain the difference between an element and a compound. Give two examples of each.

Q17. Write the chemical formulae of the following compounds:

a) Magnesium chloride
b) Calcium oxide
c) Aluminium oxide

Q18. Calculate the relative molecular mass of water, H₂O, if: H=1,O=16

Q19. Explain how the formula of magnesium chloride, MgCl₂, is obtained using the valencies of magnesium and chlorine.

Section D — Application-Based Questions

2 × 4 = 8 Marks

Q20. A student has to write the formula of aluminium oxide. Aluminium has valency 3 and oxygen has valency 2.

Answer the following:

a) Write the valency of aluminium.
b) Write the valency of oxygen.
c) Use the criss-cross method to determine the formula.
d) Why is the final formula written in the simplest whole-number ratio?

Q21. Consider the following chemical formula: Ca(OH)2​

Answer the following:

a) Name the compound.
b) How many calcium atoms are present in one formula unit?
c) How many oxygen atoms are present?
d) How many hydrogen atoms are present?

Section E — Long Answer and Numerical Questions

2 × 5 = 10 Marks

Q22. Explain the main ideas of the atomic theory proposed by John Dalton. Mention any four important points.

Q23. Calculate the relative molecular masses of the following substances:

a) CO₂
b) NH₃
c) CaCO₃

Use: C=12,O=16,N=14,H=1,Ca=40

SOLUTIONS

Section A — Answers

Q1. c) Is neither created nor destroyed in a chemical reaction

Q2. b) Atom

Q3. b) O₂

Q4. a) H₂O

Q5. b) Valency

Q6. c) Water

Q7. a) NaCl

Q8. b) Cation

Q9. a) Adding the atomic masses of all atoms in its molecule

Q10. a) The same elements in the same fixed proportion by mass

Section B — Solutions

Q11. Law of Conservation of Mass

The law of conservation of mass states that mass can neither be created nor destroyed during a chemical reaction.

Therefore: Total mass of reactants = Total mass of products​

provided the reaction is considered in a closed system.

Q12. Atom and Molecule

An atom is the smallest unit of an element that retains the chemical identity of that element.

A molecule is a group of two or more atoms chemically bonded together that can exist as an independent unit.

Examples:

• Oxygen atom = O
• Oxygen molecule = O₂
• Water molecule = H₂O

Q13. Valency

Valency is the combining capacity of an atom or element.

In water: H2​O

Hydrogen has valency 1 and oxygen has valency: 2​

Thus, the valency of oxygen is 2.

Q14. Cation and Anion

Q15. Law of Definite Proportions

The law of definite proportions states that a pure chemical compound always contains the same elements combined in the same fixed proportion by mass.

For example, pure water always contains hydrogen and oxygen in a fixed mass ratio.

Thus, the composition of a pure compound is definite and does not depend on its source.

Section C — Solutions

Q16. Element and Compound

An element is a pure substance consisting of only one kind of atom.

Examples:

• Iron (Fe)
• Oxygen (O)

A compound is a pure substance formed when two or more different elements combine chemically in a fixed proportion.

Examples:

• Water (H₂O)
• Carbon dioxide (CO₂)

Q17. Chemical Formulae

a) Magnesium chloride

Magnesium valency = 2

Chlorine valency = 1

Therefore: MgCl2​​

b) Calcium oxide

Calcium valency = 2

Oxygen valency = 2

The ratio simplifies to 1:1: CaO​

c) Aluminium oxide

Aluminium valency = 3

Oxygen valency = 2

Using the criss-cross method: Al3+, O2−

Therefore: Al2​O3​​

Q18. Relative Molecular Mass of Water

Formula: H2​O

Given: H=1 O=16

Therefore: Mr​=(2×1)+16 Mr​=2+16 Mr​=18​

Answer: Relative molecular mass of water = 18.

Q19. Formula of Magnesium Chloride

Valency of magnesium: Mg=2

Valency of chlorine: Cl=1

We need two chloride ions for every magnesium ion to balance the charges. Mg2++2Cl−→MgCl2​

Therefore, the formula is: MgCl2​​

The total positive and negative charges are balanced: +2+(2×−1)=0

Section D — Solutions

Q20. Aluminium Oxide

Given:

Valency of aluminium = 3

Valency of oxygen = 2

a) Aluminium valency

3​

b) Oxygen valency

2​

c) Formula using criss-cross method

Write: Al3+O2−

Cross the valencies: Al2​O3​

Therefore: Al2​O3​​

d) Simplest whole-number ratio

Chemical formulae are written using the simplest whole-number ratio of the atoms or ions present.

For aluminium oxide, the ratio is: Al:O=2:3

Therefore, the correct formula is: Al2​O3​​

Q21. Calcium Hydroxide

Given formula: Ca(OH)2​

a) Name

Calcium hydroxide​

b) Calcium atoms

There is no subscript after Ca.

Therefore: 1 calcium atom​

c) Oxygen atoms

The group OH occurs twice: (OH)2​

Therefore: 2 oxygen atoms​

d) Hydrogen atoms

The OH group occurs twice, so: 2 hydrogen atoms​

Thus, one formula unit contains:

• 1 Ca atom
• 2 O atoms
• 2 H atoms

Section E — Solutions

Q22. Dalton’s Atomic Theory

John Dalton proposed an atomic theory to explain the nature of matter and chemical combination.

Important points include:

1. Matter consists of atoms

All matter is composed of very small particles called atoms.

2. Atoms of an element

Atoms of the same element were proposed to have the same basic properties and mass.

3. Different elements

Atoms of different elements differ in their properties and masses.

4. Formation of compounds

Atoms of different elements combine in simple whole-number ratios to form compounds.

5. Chemical reactions

In a chemical reaction, atoms are rearranged. They are not created or destroyed.

Dalton’s theory provided an important foundation for understanding chemical combination and the composition of matter.

Q23. Relative Molecular Masses

Given: C=12,O=16,N=14,H=1,Ca=40

a) CO₂

Formula: CO2​

Therefore: Mr​=12+(2×16) Mr​=12+32 Mr​=44​

b) NH₃

Formula: NH3​

Therefore: Mr​=14+(3×1) Mr​=14+3 Mr​=17​

c) CaCO₃

Formula: CaCO3​

Therefore: Mr​=40+12+(3×16) Mr​=40+12+48 Mr​=100​

Final Answers