Atomic Foundations of Matter — Class 9 Science NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 9 Science chapter "Atomic Foundations of Matter" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 9 Science chapter "Atomic Foundations of Matter" — 8 important questions with detailed answers for CBSE bo…
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NCERT Solutions for Class 9 Science — Atomic Foundations of Matter. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Atomic Foundations of Matter — All 8 Questions Solved
Q1. State and explain the Law of Conservation of Mass.
Law of Conservation of Mass:
Mass cannot be created or destroyed in a chemical reaction; mass of reactants = mass of products.
Explanation: In chemical reaction, bonds break and reform between atoms. Atoms rearrange but total number of each type of atom remains same, so total mass remains constant.
Example: Hydrogen + Oxygen → Water
2H₂ + O₂ → 2H₂O
4 g + 32 g → 36 g (mass conserved)
Q2. State and explain the Law of Definite Proportions.
Law of Definite Proportions (Law of Constant Composition):
A compound always contains the same elements in the same proportion by mass, regardless of source or method of preparation.
Example: Water always contains H and O in ratio 1:8 by mass
2 g H : 16 g O in any water sample (H₂O)
Explanation: A compound has fixed chemical formula determined by specific number of atoms of each element; ratio of atoms = ratio of masses.
Q3. Define atomic mass unit (u) and atomic mass.
Atomic mass unit (u or amu):
Standard unit for measuring mass of atoms and molecules.
1 u = 1/12 of mass of one carbon-12 atom = 1.66 × 10⁻²⁷ kg
Atomic mass = Average relative mass of atom of an element compared to 1/12 of C-12
Example: Atomic mass of Hydrogen ≈ 1 u, Carbon = 12 u, Oxygen = 16 u
Molecular mass = Sum of atomic masses of all atoms in molecule
Example: H₂O = 1 + 1 + 16 = 18 u
Q4. What is the mole concept? Define mole and molar mass.
Mole = Unit of amount of substance; 1 mole = 6.022 × 10²³ particles (Avogadro's number Nₐ)
Molar mass (M) = Mass of one mole of substance in grams = Numerically equal to molecular/atomic mass
Example: Molar mass of H₂O = 18 g/mol, O₂ = 32 g/mol
Relationship: n = m/M = N/Nₐ
where n = number of moles, m = mass (g), N = number of particles
Applications: Convert between mass, moles, and number of particles
Q5. Calculate the number of moles in 36 g of water (H₂O).
Given: mass of water = 36 g
Molar mass of H₂O = 2(1) + 16 = 18 g/mol
Using formula: n = m/M
n = 36/18 = 2 moles
Alternatively: Number of molecules = n × Nₐ = 2 × 6.022 × 10²³ = 1.204 × 10²⁴ molecules
Q6. What is a chemical formula? Explain empirical and molecular formula with example.
Chemical formula = Symbolic representation of compound showing types and relative numbers of atoms of each element.
Empirical formula = Simplest whole number ratio of atoms of different elements
Molecular formula = Actual number of atoms of each element in one molecule
Example: Glucose C₆H₁₂O₆
Empirical formula = CH₂O (ratio 1:2:1)
Molecular formula = C₆H₁₂O₆ (actual number of atoms)
Relationship: Molecular formula = (Empirical formula)ₙ
Q7. How many atoms are present in 2 moles of CO₂?
Given: 2 moles of CO₂
One molecule of CO₂ contains = 1 C atom + 2 O atoms = 3 atoms total
Number of molecules in 2 moles = 2 × Nₐ = 2 × 6.022 × 10²³ = 1.204 × 10²⁴ molecules
Total atoms = 3 × 1.204 × 10²⁴ = 3.612 × 10²⁴ atoms
Alternatively: 2 moles × 3 atoms/molecule × 6.022 × 10²³ = 3.612 × 10²⁴ atoms
Q8. A compound contains C, H, O in ratio 1:2:1 by atoms. If its molar mass is 60 g/mol, find molecular formula.
Empirical formula from ratio 1:2:1 = CH₂O
Empirical mass of CH₂O = 12 + 2(1) + 16 = 30 u
n = Molar mass / Empirical mass = 60/30 = 2
Molecular formula = (CH₂O)₂ = C₂H₄O₂
This is acetic acid (ethanoic acid)
Molecular mass = 2(12) + 4(1) + 2(16) = 24 + 4 + 32 = 60 g/mol ✓
Atomic Foundations of Matter — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. According to the Law of Conservation of Mass, what happens to the total mass of reactants and products during a chemical reaction?
- A. It increases.
- B. It decreases.
- C. It remains constant.
- D. It first increases then decreases.
Answer: C. It remains constant. — The Law of Conservation of Mass states that mass can neither be created nor destroyed in a chemical reaction. Therefore, the total mass of reactants must equal the total mass of products.
MCQ 2. If 10 g of calcium carbonate is decomposed, it yields 4.4 g of carbon dioxide and 'x' g of calcium oxide. What is the value of 'x' according to the Law of Conservation of Mass?
- A. 14.4 g
- B. 5.6 g
- C. 4.4 g
- D. 10 g
Answer: B. 5.6 g — According to the Law of Conservation of Mass, the total mass of reactants (10 g) must be equal to the total mass of products (4.4 g + x). So, 10 = 4.4 + x, which gives x = 5.6 g.
MCQ 3. Which law states that in a chemical substance, the elements are always present in definite proportions by mass?
- A. Law of Conservation of Mass
- B. Law of Multiple Proportions
- C. Law of Constant Proportions
- D. Law of Combining Volumes
Answer: C. Law of Constant Proportions — The Law of Constant Proportions (also known as the Law of Definite Proportions) states that a chemical compound always contains exactly the same proportion of elements by mass, regardless of the source.
MCQ 4. Which of the following is NOT a postulate of Dalton's Atomic Theory?
- A. All matter is made of tiny, indivisible particles called atoms.
- B. Atoms of a given element are identical in all respects.
- C. Atoms can be divided into protons, neutrons, and electrons.
- D. Atoms combine in simple whole-number ratios to form compounds.
Answer: C. Atoms can be divided into protons, neutrons, and electrons. — Dalton's Atomic Theory stated that atoms are indivisible particles. The discovery of subatomic particles later proved this postulate incorrect.
MCQ 5. What is the atomic mass unit (amu) now commonly referred to as?
- A. Gram
- B. Unified atomic mass unit (u)
- C. Kilogram
- D. Mole
Answer: B. Unified atomic mass unit (u) — The atomic mass unit (amu) is now commonly referred to as the unified atomic mass unit (u), which is defined as one-twelfth the mass of an uncharged atom of carbon-12.
MCQ 6. What is the valency of the sulphate ion (SO4^2-)?
- A. 1
- B. 2
- C. 3
- D. 4
Answer: B. 2 — The valency of a polyatomic ion is equal to the magnitude of its charge. The sulphate ion has a charge of -2, so its valency is 2.
MCQ 7. What is the chemical formula for Magnesium Chloride? (Valency: Mg=2, Cl=1)
- A. MgCl
- B. MgCl2
- C. Mg2Cl
- D. Mg2Cl2
Answer: B. MgCl2 — Magnesium has a valency of 2 and Chlorine has a valency of 1. To balance the valencies, one magnesium atom combines with two chlorine atoms, resulting in the formula MgCl2.
MCQ 8. Calculate the molecular mass of H2O. (Atomic mass: H=1 u, O=16 u)
- A. 17 u
- B. 18 u
- C. 19 u
- D. 16 u
Answer: B. 18 u — The molecular mass of H2O is calculated by adding the atomic masses of its constituent atoms: (2 × mass of H) + (1 × mass of O) = (2 × 1 u) + (1 × 16 u) = 2 u + 16 u = 18 u.
MCQ 9. One mole of any substance contains how many particles (atoms, molecules, or ions)?
- A. 6.022 x 10^22
- B. 6.022 x 10^23
- C. 6.022 x 10^24
- D. 1 x 10^23
Answer: B. 6.022 x 10^23 — One mole of any substance contains Avogadro's number of particles, which is approximately 6.022 x 10^23 particles.
MCQ 10. What is the molar mass of Carbon Dioxide (CO2)? (Atomic mass: C=12 u, O=16 u)
- A. 28 g/mol
- B. 32 g/mol
- C. 44 g/mol
- D. 12 g/mol
Answer: C. 44 g/mol — Molar mass of CO2 = (1 × mass of C) + (2 × mass of O) = (1 × 12 g/mol) + (2 × 16 g/mol) = 12 + 32 = 44 g/mol.
MCQ 11. How many moles are present in 23 g of Sodium (Na)? (Atomic mass of Na = 23 g/mol)
- A. 0.5 mol
- B. 1 mol
- C. 2 mol
- D. 23 mol
Answer: B. 1 mol — Number of moles = Given mass / Molar mass. Here, 23 g / 23 g/mol = 1 mole.
MCQ 12. Which scientist proposed the atomic theory that stated atoms are indivisible and indestructible particles?
- A. Antoine Lavoisier
- B. Joseph Proust
- C. John Dalton
- D. J.J. Thomson
Answer: C. John Dalton — John Dalton proposed the first comprehensive atomic theory, which included the postulate that atoms are indivisible and indestructible.
MCQ 13. A sample of ammonia (NH3) always contains Nitrogen and Hydrogen in the ratio 14:3 by mass. This is an example of the:
- A. Law of Conservation of Mass
- B. Law of Multiple Proportions
- C. Law of Constant Proportions
- D. Law of Gaseous Volumes
Answer: C. Law of Constant Proportions — The fixed mass ratio of nitrogen to hydrogen in ammonia illustrates the Law of Constant Proportions, stating that a pure chemical compound always contains elements in fixed mass ratios.
MCQ 14. What is the atomicity of ozone (O3)?
- A. 1
- B. 2
- C. 3
- D. 4
Answer: C. 3 — Atomicity is the number of atoms present in a molecule. Ozone (O3) consists of three oxygen atoms, so its atomicity is 3.
MCQ 15. What is the chemical formula for Calcium Hydroxide? (Valency: Ca=2, OH=1)
- A. CaOH
- B. Ca(OH)2
- C. Ca2OH
- D. Ca(OH)
Answer: B. Ca(OH)2 — Calcium has a valency of 2 and the hydroxide ion (OH) has a valency of 1. To balance the valencies, one calcium atom combines with two hydroxide ions, resulting in the formula Ca(OH)2.
MCQ 16. How many moles are present in 12.044 x 10^23 atoms of Helium?
- A. 0.5 mol
- B. 1 mol
- C. 2 mol
- D. 4 mol
Answer: C. 2 mol — One mole contains 6.022 x 10^23 particles (Avogadro's number). Therefore, 12.044 x 10^23 atoms correspond to (12.044 x 10^23) / (6.022 x 10^23) = 2 moles.
MCQ 17. What is the formula unit mass of Sodium Chloride (NaCl)? (Atomic mass: Na=23 u, Cl=35.5 u)
- A. 58.5 u
- B. 23 u
- C. 35.5 u
- D. 1 u
Answer: A. 58.5 u — The formula unit mass of NaCl is the sum of the atomic masses of Na and Cl: 23 u + 35.5 u = 58.5 u.
MCQ 18. A sample of carbon burning in oxygen produces 4.4 g of carbon dioxide. If 1.2 g of carbon was used, how much oxygen was consumed?
- A. 3.2 g
- B. 5.6 g
- C. 4.4 g
- D. 1.2 g
Answer: A. 3.2 g — According to the Law of Conservation of Mass, Mass(Carbon) + Mass(Oxygen) = Mass(Carbon Dioxide). So, 1.2 g + Mass(Oxygen) = 4.4 g. Therefore, Mass(Oxygen) = 4.4 - 1.2 = 3.2 g.
MCQ 19. How many atoms of oxygen are present in 0.5 moles of Carbon Dioxide (CO2)? (Avogadro number = 6.022 x 10^23)
- A. 6.022 x 10^23 atoms
- B. 1.2044 x 10^24 atoms
- C. 3.011 x 10^23 atoms
- D. 1.2044 x 10^23 atoms
Answer: A. 6.022 x 10^23 atoms — In one molecule of CO2, there are 2 oxygen atoms. In 0.5 moles of CO2, there are 0.5 × (6.022 x 10^23) molecules. Therefore, the number of oxygen atoms = 2 × 0.5 × (6.022 x 10^23) = 6.022 x 10^23 atoms.
MCQ 20. Which of the following polyatomic ions has a valency of 3?
- A. Nitrate (NO3-)
- B. Sulphate (SO4^2-)
- C. Phosphate (PO4^3-)
- D. Carbonate (CO3^2-)
Answer: C. Phosphate (PO4^3-) — The valency of a polyatomic ion is given by the magnitude of its charge. Phosphate (PO4^3-) has a charge of -3, hence its valency is 3.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.