Class 9 Science Chapter 3 question answer Atoms and Molecules

Ncert Solutions for Class 9 Science Chapter 3: Atoms and Molecules class 9 questions and answers

Textbook	Ncert
Class	Class 9
Subject	Science
Chapter	Chapter 3
Chapter Name	Atoms and Molecules class 9 ncert solutions
Category	Ncert Solutions
Medium	English

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In Text Questions Page No: 27

Question 1: In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.
Sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water

Answer 1: To show that the observations agree with the law of conservation of mass, we need to check whether the total mass of the reactants equals the total mass of the products.

Add the masses of the reactants

• Sodium carbonate = 5.3 g
• Ethanoic acid = 6.0 g

Total mass of reactants = 5.3 g + 6.0 g = 11.3 g

Add the masses of the products

• Carbon dioxide = 2.2 g
• Water = 0.9 g
• Sodium ethanoate = 8.2 g

Total mass of products = 2.2 g + 0.9 g + 8.2 g = 11.3 g

Compare the two totals

• Reactants: 11.3 g
• Products: 11.3 g

Since the mass of the reactants equals the mass of the products, the data agrees with the law of conservation of mass.

Question 2: Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?

Answer 2: we’ll use the mass ratio given in the question: Hydrogen : Oxygen = 1 : 8 by mass

Set up the ratio

If 1 g of hydrogen reacts with 8 g of oxygen,
then 3 g of hydrogen will react with: 3×8=24 g of oxygen

24 g of oxygen gas would be required to react completely with 3 g of hydrogen gas.

Question 3: Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?

Answer 3: The postulate of Dalton :”Atoms are indivisible particles, which can not be created or destroyed in a chemical reaction” is the result of the law of conservation of mass.

Question 4: Which postulate of Dalton’s atomic theory can explain the law of definite proportions?

Answer 4: The postulate of Dalton’s atomic theory that can explain the law of definite proportions is that the relative number and kinds of atoms are equal in given compounds.

In Text Questions Page No: 30

Question 1: Define the atomic mass unit.

Answer 1: Definition of Atomic Mass Unit (amu): One atomic mass unit (1 amu) is defined as one-twelfth the mass of a carbon-12 atom.

In other words:1 amu = \(\frac{1}{12}\) the mass of one atom of carbon-12
≈ 1.66 × 10⁻²⁷ kg

This unit is used to express the masses of atoms and molecules in a manageable way, since their actual masses in kilograms are extremely small.

Question 2: Why is it not possible to see an atom with naked eyes?

Answer 2: Due to small size of an atom we cannot see them with naked eyes.

In Text Questions Page No: 34

Question 1: Write down the formulae of
(i) sodium oxide
(ii) aluminium chloride
(iii) sodium sulphide
(iv) magnesium hydroxide

Answer 1:

• (i) Sodium oxide → Na₂O
• (ii) Aluminium chloride → AlCl₃
• (iii) Sodium sulphide → Na₂S
• (iv) Magnesium hydroxide → Mg(OH)₂

Question 2: Write down the names of the compounds represented by the following formulae:
(i) Al₂(SO₄)₃
(ii) CaCl₂
(iii) K₂SO₄
(iv) KNO₃
(v) CaCO₃

Answer 2: Listed below are the names of the compounds for each of the following formulae:

• (i) Al₂(SO₄)₃ – Aluminium sulphate
• (ii) CaCl₂ – Calcium chloride
• (iii) K₂SO₄ – Potassium sulphate
• (iv) KNO₃ – Potassium nitrate
• (v) CaCO₃ – Calcium carbonate

Question 3: What is meant by the term chemical formula?

Answer 3: The chemical formula of a compound means the symbolic representation of the composition of a compound.

Question 4: How many atoms are present in a
(i) H₂S molecule and
(ii) PO₄^3- ion?

Answer 4: The number of atoms present is as follows: (i) H₂S molecule has 2 atoms of hydrogen and 1 atom of sulphur hence 3 atoms in total. (ii) PO₄^3- ion has 1 atom of phosphorus and 4 atoms of oxygen hence 5 atoms in total.

In Text Questions Page No: 35

Question 1: Calculate the molecular masses of H₂, O₂, Cl₂, CO₂, CH₄, C₂H₆, C₂H₄, NH₃, CH₃OH.

Answer 1: Molecular Masses:

1. H₂ = 2 × 1 = 2 g/mol
2. O₂ = 2 × 16 = 32 g/mol
3. Cl₂ = 2 × 35.5 = 71 g/mol
4. CO₂ = 12 + (2 × 16) = 12 + 32 = 44 g/mol
5. CH₄ (methane) = 12 + (4 × 1) = 12 + 4 = 16 g/mol
6. C₂H₆ (ethane) = (2 × 12) + (6 × 1) = 24 + 6 = 30 g/mol
7. C₂H₄ (ethene) = (2 × 12) + (4 × 1) = 24 + 4 = 28 g/mol
8. NH₃ (ammonia) = 14 + (3 × 1) = 14 + 3 = 17 g/mol
9. CH₃OH (methanol) = 12 + (4 × 1) + 16 = 12 + 4 + 16 = 32 g/mol

Question 2: Calculate the formula unit masses of ZnO, Na₂O, K₂CO₃, given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.

Answer 2:

(i) ZnO

• Zn = 65 u
• O = 16 u
  Formula unit mass = 65 + 16 = 81 u

(ii) Na₂O

• Na = 23 u (×2) = 46 u
• O = 16 u
  Formula unit mass = 46 + 16 = 62 u

(iii) K₂CO₃

• K = 39 u (×2) = 78 u
• C = 12 u
• O = 16 u (×3) = 48 u
  Formula unit mass = 78 + 12 + 48 = 138 u

Exercises

Question 1: A 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.

Answer 1: Given: Mass of the sample compound = 0.24g, mass of boron = 0.096g, mass of oxygen = 0.144g

To calculate the percentage composition of the compound,

Percentage of boron = mass of boron / mass of the compound x 100

= 0.096g / 0.24g x 100  = 40%

Percentage of oxygen = 100 – percentage of boron

= 100 – 40 = 60%

Question 2: When 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?

Answer 2: Carbon + Oxygen ⎯⎯→ Carbon dioxide 

3 g of carbon reacts with 8 g of oxygen to produce 11 g of carbon dioxide. 

If 3 g of carbon is burnt in 50 g of oxygen, then 3 g of carbon will react with 8 g of oxygen to form11 g of carbon dioxide. 

The remaining (50 –8) = 42 g of oxygen will be left unreacted. 

The above answer is governed by the law of constant proportions.

Question 3: What are polyatomic ions? Give examples.

Answer 3: Polyatomic ions are ions that contain more than one atom, but they behave as a single unit. Example: CO₃^2-, H₂PO₄^–

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

Answer 4: The following are the chemical formula of the above-mentioned list:

• (a) Magnesium chloride – MgCl₂
• (b) Calcium oxide – CaO
• (c) Copper nitrate – Cu(NO₃)₂
• (d) Aluminium chloride – AlCl₃
• (e) Calcium carbonate – CaCO₃

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

Answer 5: (a) Quick lime (CaO): The elements present in it are calcium and oxygen.
(b) Hydrogen bromide (HBr): The elements present in it are hydrogen and bromine.
(c) Baking powder (NaHCO₃): The elements present in it are sodium, hydrogen, carbon and oxygen.
(d) Potassium sulphate (K₂SO₄): The elements present in it are potassium, sulphur and oxygen.

Question 6: Calculate the molar mass of the following substances:
(a) Ethyne, C₂H₂ (b) Sulphur molecule, S₈
(c) Phosphorus molecule, P₄ (Atomic mass of phosphorus is 31)
(d) Hydrochloric acid, HCl (e) Nitric acid, HNO₃

Answer 6: Listed below is the molar mass of the following substances:

• (a) Ethyne (C₂H₂) = (2 × 12) + (2 × 1) = 24 + 2 = 26 g/mol
• (b) Sulphur molecule (S₈) = 8 × 32 = 256 g/mol
• (c) Phosphorus molecule (P₄) = 4 × 31 = 124 g/mol
• (d) Hydrochloric acid (HCl) = 1 + 35.5 = 36.5 g/mol
• (e) Nitric acid (HNO₃) = 1 + 14 + (3 × 16) = 1 + 14 + 48 = 63 g/mol