Atoms and Molecules Science Class 9 Chapter 3 Question Answer

Solutions For All Chapters Science Class 9

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 polyatomic 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., 

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

(e) 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

NCERT Textbook for Class 9 Science – Page 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 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 —> 
∴ 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 30

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 Textbook Questions – Page 34

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

(i) Al₂(SO₄)₃

(ii) CaCl₂

(iii) K₂SO₄

(iv) KNO₃

(v) CaCO₃.

Solution:

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: The chemical formula of the compound is a symbolic representation of its composition, e.g., chemical formula of sodium chloride is NaCl.

Question 4. How many atoms are present in a
(i) H2S molecule and
(ii) P043- ion?

Answer: (i) H2S —> 3 atoms are present
(ii) P043- —> 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