Class 9 Science Ch 3 Atoms and Molecules – NCERT Answers

This chapter introduces the fundamental concepts of chemistry by explaining the basic building blocks of matter: atoms and molecules. It covers the laws of chemical combination, such as the Law of Conservation of Mass and the Law of Constant Proportions. Students learn about atomic mass, molecular mass, and how to calculate them. The chapter also explains the concept of moles, chemical formulae, and how to write and balance chemical equations. It lays the foundation for understanding how substances interact and form new compounds.

Class 9 Science Ch 3 Atoms and Molecules – Textbook

Page 32
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 carbonate.
Answer.

TEXT QUESTIONS SOLVED

Class 9 Science NCERT Textbook Page 32
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 carbonate.
Answer:

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: Ratio of H : O by mass in water is:
Hydrogen : Oxygen —> H₂O
∴ 1 : 8 = 3 : x
x = 8 x 3
x = 24 g
∴ 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: The postulate of Dalton’s atomic theory that is the result of the law of conservation of mass is—the relative number and kinds of atoms are constant in a given compound. Atoms cannot be created nor destroyed in a chemical reaction.

Question 4. Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Answer: The relative number and kinds of atoms are constant in a given compound.

Class 9 Science NCERT Textbook Page 35
Question 1. Define the atomic mass unit.
Answer: One atomic mass unit is equal to exactly one-twelfth (1/12th) the mass of one atom of carbon-12. The relative atomic masses of all elements have been found with respect to an atom of carbon-12.

Question 2. Why is it not possible to see an atom with naked eyes?
Answer: Atom is too small to be seen with naked eyes. It is measured in nanometres.
1 m = 109 nm

NCERT Solutions For Class 9 Science Chapter 3 Atoms and Molecules

NCERT Textbook Questions – Page 39
Question 1. Write down the formulae of
(i) Sodium oxide
(ii) Aluminium chloride
(iii) Sodium sulphide
(iv) Magnesium hydroxide
Answer: The formulae are

Question 2. What is meant by the term chemical formula?
Answer: The chemical formula of the compound is a symbolic representation of its composition, e.g., chemical formula of sodium chloride is NaCl.

Question 3. How many atoms are present in a
(i) H₂S molecule and
(ii) P0₄^3- ion?
Answer: (i) H₂S —> 3 atoms are present
(ii) P0₄^3- —> 5 atoms are present

NCERT Textbook Questions – Page 40
Question 1. Calculate the molecular masses of H₂, O₂, Cl₂, C0₂, CH₄, C₂H₂,NH₃, CH₃OH.
Answer: The molecular masses are:

Question 2.Calculate the formula unit masses of ZnO, Na₂O, K₂C0₃, given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.
Answer: The formula unit mass of
(i) ZnO = 65 u + 16 u = 81 u
(ii) Na₂O = (23 u x 2) + 16 u = 46 u + 16 u = 62 u
(iii) K₂C0₃ = (39 u x 2) + 12 u + 16 u x 3
= 78 u + 12 u + 48 u = 138 u

Questions From NCERT Textbook for Class 9 Science

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: Boron and oxygen compound —> Boron + Oxygen
0.24 g —> 0.096 g + 0.144 g

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: The reaction of burning of carbon in oxygen may be written as:

It shows that 12 g of carbon bums in 32 g oxygen to form 44 g of carbon dioxide. Therefore 3 g of carbon reacts with 8 g of oxygen to form 11 g of carbon dioxide. It is given that 3.0 g of carbon is burnt with 8 g of oxygen to produce 11.0 g of CO₂. Consequently 11.0 g of carbon dioxide will be formed when 3.0 g of C is burnt in 50 g of oxygen consuming 8 g of oxygen, leaving behind 50 – 8 = 42 g of O₂. The answer governs the law of constant proportion.

Question 3. What are poly atomic ions? Give examples.
Answer: The ions which contain more than one atoms (same kind or may be of different kind) and behave as a single unit are called polyatomic ions e.g., OH^–, SO₄^2-, CO₃^2-.

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: (a) Magnesium chloride
Symbol —> Mg Cl
Change —> +2 -1
Formula —> MgCl₂
(b) Calcium oxide
Symbol —> Ca O
Charge —> +2 -2
Formula —> CaO
(c) Copper nitrate
Symbol —> Cu NO
Change +2 -1
Formula -4 CU(N0₃)₂
(d) Aluminium chloride
Symbol —> Al Cl
Change —> +3 -1
Formula —> AlCl₃
(d) Calcium carbonate
Symbol —> Ca CO₃
Change —> +2 -2
Formula —> CaC0₃

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: (a) Quick lime —> Calcium oxide
Elements —> Calcium and oxygen
(b) Hydrogen bromide
Elements —> Hydrogen and bromine
(c) Baking powder —> Sodium hydrogen carbonate
Elements —> Sodium, hydrogen, carbon and oxygen
(d) Potassium sulphate
Elements —> 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 = 31)
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO₃
Answer: The molar mass of the following: [Unit is ‘g’]
(a) Ethyne, C₂H₂ = 2 x 12 + 2 x 1 = 24 + 2 = 26 g
(b) Sulphur molecule, S₈ = 8 x 32 = 256 g
(c) Phosphorus molecule, P₄=4 x 31 = i24g
(d) Hydrochloric acid, HCl = 1 x 1 + 1 x 35.5 = 1 + 35.5 = 36.5 g
(e) Nitric acid, HN0₃ = 1 x 1 + 1 x 14 + 3 x 16 = 1 + 14 + 48 = 63 g

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NCERT Class 10 Mathematics Solutions

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