Class 9 Science Chapter 3 Atoms and Molecules Solutions | ASSEB Assam | English Medium

Class 9 Science Chapter 3 – Atoms and Molecules – All Textual Solutions | ASSEB Assam (English Medium)

Class 9 Science Chapter 3 – Atoms and Molecules Complete Textual Questions and Solutions (ASSEB / SEBA Assam – English Medium)

The third chapter of the Class 9 Science (English Medium) syllabus, ‘Atoms and Molecules’, is a foundational chapter that introduces students to the basic building blocks of matter and the rules governing chemical reactions. Based on the latest guidelines of the new ASSEB (Assam State School Education Board) syllabus and the New Education Policy (NEP), this comprehensive textual question-answer guide has been prepared for upcoming board examinations. This exclusive collection includes Very Short Answers (VSA), Short Questions, Long Answers, and step-by-step numerical solutions. Ospin Academy presents these solutions in an easy, 100% accurate, and completely exam-oriented format.

In this chapter, ‘Atoms and Molecules’, students explore the laws of chemical combination, Dalton’s atomic theory, and the concepts of atoms, molecules, and ions. A significant portion of the chapter focuses on understanding valency and writing chemical formulae correctly, as well as calculating molecular mass. For students preparing for their exams, mastering these chemical formulas and numericals is essential. Our textual solutions cover not only the textbook exercise questions but also additional crucial questions that frequently appear in the board exams, ensuring students are thoroughly prepared.

What you will learn and get from these Textual Solutions:

• Detailed explanations of the Laws of Chemical Combination and Dalton’s Atomic Theory.
• Flawless answers for 1-mark Very Short Answers (VSA objective type) and 2-3 mark conceptual questions.
• Step-by-step, simple methods for writing chemical formulae of various compounds.
• Clear, accurate solutions for numerical problems based on calculating molecular mass and formula unit mass.
• Important questions and the latest pattern of MCQ solutions according to the new syllabus format.

Special Features of Ospin Academy’s Question & Answers:

• Complete textual solutions prepared strictly according to the current ASSEB new syllabus.
• 100% accurate, high-quality notes written in simple English, making it easy for Class 9 students to understand and practice.
• Simplified answers and numerical breakdowns specifically curated for quick revision before exams.
• An exclusive collection of extra important questions that carry a high probability of appearing in the exams.

Keeping the academic success of students in mind, Ospin Academy has provided these special textual solutions. Start your exam preparation today and step forward towards scoring the highest marks in your Science examinations.

Class 9 Science

Chapter – 3 ‌ (Ospin Academy)

Atoms and Molecules

Textbook Page No. 32-33

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 log → sodium ethanoate + carbon dioxide + water

Ans. Total massa of the reactants

= Mass of sodium carbonate+Mass of ethanoic acid

= 5.3g+6g

= 11.3g

Total mass of the products

= Mass of carbon dioxide + Mass of water + Mass of sodium ethanoate

= 2.2g + 0.9g + 8.2g

= 11.2g

Since the total mass of the reactants is equal to the total mass of the products, these observations are in agreement with the law of conservation of mass.

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 3g of hydrogen gas?

Ans. 1 g of hydrogen reacts with 8 gm of oxygen to form water. Therefore, according to the law of constant proportions, Mass of oxygen gas required to react completely with 3 g of hydrogen gas is 8 × 3g = 24 g

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

Ans. The postulate of Dalton’s atomic theory is the results of the law of conversation of mass are:

(i) All matter is composed of the indivisible particles called atoms. Atoms of an element cannot be created neither can be destroyed.

(ii) Atoms of the same element have similar shape and mass, but the mass and shape can vary for atoms of other elements.

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

Ans. The postulate of Dalton’s atomic theory that can explain the law of definite proportions is- “The relative mumber and kinds of atoms are constant in a given compound.”

Textbook Page No. 35

1. Define the atomic unit.

Ans. The atomic mass unit (amu) is the quantity of mass equal to exactly one-twelfth (photo) the mass of one atom of C¹²

One atomic mass unit (1u)

1 u = (1/12) × mass of one C¹² atom

Numerically,

1 u = 1.66 × 10⁻²⁷ kg

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

Ans. An atom is considered to be the smallest unit of a matter. Its size is smaller than anything that we can imagine or compare with. If more than millions of atoms are stacked, they will make a layer barely as thick as a sheet of paper. Hence, it is not possible to see an atom with naked eyes.

Textbook Page No. 39

1. Write down formula of-

(i) Sodium oxide.

Ans. (i) Na₂O.

(ii) Aluminium chloride.

Ans: AICI₃.

(iii) Sodium sulphide.

 Ans: Na₂S.

(iv) Magnesium hydroxide.

Ans: Mg(OH)₂.

2. Write down the names of compounds represented by the following formulae:

(i) AI₂(SO⁴)₃.

Ans. (i) Aluminium sulphate.

(ii) CaCI₂.

Ans. Calcium chloride.

(iii) K₂SO₄.

Ans. Potassium sulphate.

(iv) KNO₃.

Ans. Potassium nitrate.

(v) CaCO₃.

Ans. Calcium carbonate.

3. What is meant by the term chemical formula?

Ans. The chemical formula of a compound is a symbolic representation of its composition. It indicates the number and types of atoms of different elements present in one molecule of the compound. For example, carbon dioxide is a compound made up of 1 atom of carbon and 2 atoms of oxygen.

Therefore, the formula of carbon dioxide is written as CO₂.

4. How many atoms are present in a

• 1. H₂S molecule.

Ans. H₂S molecule: 3 atoms; 2 atoms of hydrogen and 1 atom of sulphur.

PO₄³⁻ ion?

Ans: PO₄³{O ion: 5 atoms; 1 atom of phosphorus and 4 atoms of oxygen.

Textbook Page No. 40

1. Calculate the molecular masses of H₂, O₂, CI₂, CO₂, CH₄, C₂H₆, C₂H₄, NH₃, CH₃OH.

Ans. Molecular mass of H₂ = 2 × atomic of H

= 2 × 1

= 2u

Molecular mass of O₂ = 2 × atomic mass of O

= 2 × 16

= 32u

Molecular mass of CI² = 2 × atomic mass of CI

= 2×35.5

= 71u

Molecular mass of CO₂ = (1×atomic mass of C) + (2 × atomic mass of oxygen)

= (1 × 12)̇ + (2 × 16)

= 12 + 32

= 44u

Molecular mass of CH₄ = (1 x atomic mass of C) + (4 x atomic mass of hydrogen)

= (1 × 12) + (4 × 1)

= 12 + 4

= 16u

Molecular mass of C2H₆ = (2x atomic mass of C) + (6 × atomic mass of H)

= (2 × 12)+(6 × 1)

= 24 + 6

= 30u

Molecular mass of C₂H₄ = (2 × atomic mass of C) + (4 x atomic mass of H)

= (2 × 12) + (4 × 1)

= 24 + 4

= 28u

Molecular mass of NH₃ = (1 × atomic mass of N) + (3 x atomic mass of H)

= (1 × 14)+(3 × 1)

= 14 + 3

= 17 u

Molecular mass of CH₃OH

= (1x atomic mass of C) + (3 x atomic mass of H)+(1x atomic mass of O) + (1x atomic mass of H)

= (1 × 12)+(31)+(1 × 16) + (1 × 1) =

=12 + 3 + 16 + 1

= 32u.

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

Ans. Formula unit mass of ZnO = (1 × atomic mass of Zn) + (1 x atomic mass of O)

= (1 × 65) + (1 × 16)

= 65+16

= 81 u

Formula unit mass of Na₂O = (2 × atomic mass of Na) + (1 × atomic mass of O) (2 × 23) + (1 × 16)

= 46 + 16

= 62 u

Formula unit mass of K₂CO₃ = (2 × atomic mass of K) + (1 × atomic mass of C) + (3

× atomic mass of O)

= (2 × 39) + (1 × 12) + (3 × 16)

= 78 + 12 + 48

= 138u.

Textbook Page No. 42.

1. If one mole of carbon atoms weighs 12 grams, what is the mass (in grams) of 1 atom of carbon?

Ans.

The mass of one mole of carbon atoms = 12 g

The number of atoms in one mole = 6.022 x 10²³ atoms

Mass of 1 atom of carbon = 1.993 × 10²³ g

Thus, the mass of 1 atom of carbon is 1.993 x 10²³ grams.

2. Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of Na= 23 u, Fe = 56 u)

Ans.

Atomic mass of Sodium (Na) = 23 u

Atomic mass of Iron (Fe) = 56 up

Avogadro’s number = 6.022 x 10^23

Calculations:

For Sodium (Na):

Total number of atoms = 4.35, moles x 6.022 x 10²³ = 2.62 x 10²⁴, atoms

For Iron (Fe):

Total number of atoms = 1.79, moles x 6.022 x 10²³ = 1.08 x 10²⁴, atoms

hence,

100 grams of sodium has more atoms (2.62 x 10²⁴) compared to 100 grams of iron (1.08 x 10²⁴).

(Textbook Page No. 43-44)

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.

Ans. Given:

Total mass of the compound = 0.24 g

Mass of Boron (B) = 0.096 g

Mass of Oxygen (O) = 0.144 g

Formula:

Percentage composition of an element =

Now,

Percentage of Boron (B):

And,

Percentage of Oxygen (O):

So,

Percentage of Boron = 40%

Percentage of Oxygen = 60%

2. When 3.0 g of carbon is burnt in 8.0 g of oxygen, 11.0 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.0 g of carbon is burnt in 50.0 g of oxygen? Which law of chemical combination will govern your answer?

Ans.

From the given data:

Mass of carbon dioxide = Mass of carbon + Mass of oxygen reacted = 3.0 g + 8.0 g = 11.0 g

If 3.0 g of carbon is burnt in 50.0 g of oxygen, only 8.0 g of oxygen will react, and the remaining oxygen will be unreacted.

The mass of carbon dioxide formed will still be 11.0 g.

Law of Chemical Combination:

The above reaction follows the Law of Conservation of Mass, which states that mass can neither be created nor destroyed in a chemical reaction.

3. What are polyatomic ions? Give examples.

Ans.

Polyatomic ions are ions that consist of two or more atoms chemically bonded together, carrying a net positive or negative charge.

Examples:

SO₄²⁻ (Sulfate ion)

NO₃⁻ (Nitrate ion)

NH₄⁺ (Ammonium ion)

4. Write the chemical formulae of the following:

Ans.

Magnesium chloride: MgCl₂

Calcium oxide: CaO

Copper nitrate: Cu(NO₃)₂

Aluminium chloride: AlCl₃

Calcium carbonate: CaCO₃

5. Give the names of the elements present in the following compounds:

Ans.

a) Quick lime (CaO): Calcium (Ca) and Oxygen (O)

b) Hydrogen bromide (HBr): Hydrogen (H) and Bromine (Br)

c) Baking powder (NaHCO₃): Sodium (Na), Hydrogen (H), Carbon (C), and Oxygen (O)

d) Potassium sulphate (K₂SO₄): Potassium (K), Sulphur (S), and Oxygen (O)

6. Calculate the molar mass of the following substances:

Ans.

a) Ethyne (C₂H₂):

Molar mass = (2 × 12) + (2 × 1) = 24 + 2 = 26 g/mol

b) Sulphur molecule (S₈):

Molar mass = 8 × 32 = 256 g/mol

c) Phosphorus molecule (P₄, atomic mass of phosphorus = 31):

Molar mass = 4 × 31 = 124 g/mol

d) Hydrochloric acid (HCl):

Molar mass = (1 × 1) + (1 × 35.5) = 36.5 g/mol

e) Nitric acid (HNO₃):

Molar mass = (1 × 1) + (1 × 14) + (3 × 16) = 63 g/mol