## NCERT Class 9 Atoms and Molecules Chap 3

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INTEXT QUESTIONS (Page No. 42)

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

Ans . 1 mole i.e., 6.02 × 1023 atoms of carbon weighs = 12g

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

Ans. 23 g of sodium has atoms = NA (Avogadro’s number)

Hence, 100g of sodium has more atoms than 100g of iron.

NCERT TEXTBOOK EXERCISE

Q. 1. A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.

Ans: Mass of the compound = 0.24 g Mass of boron = 0.096 g Mass of oxygen = 0.144 g

Q. 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?

Carbon + Oxygen ® Carbon dioxide

3.0 g          8.00 g             11.00 g

3.0 g carbon + 8 g oxygen = 11 g CO2 If 50 g of oxygen is there, only 8 g of O2 will react with 3.0 g of carbon. (50 – 8) = 42 g of oxygen will remain unreacted. The above data tells that the reaction between carbon and oxygen is governed by law of constant proportions.

Q. 3. What are poly atomic ions? Give examples?

Q. 3. What are poly atomic ions ? Give examples ?

Ans. The ions which contain more than one atom and behave as a single unit are called polyatomic ions. Example : Ammonium ions (NH4+ ), carbonate ion (CO3 2–).

Q. 4. Write the chemical formulae of the following:

(a) Magnesium chloride (b) Calcium oxide (c) Copper nitrate

(d) Aluminium chloride (e) Calcium carbonate.

Ans. (a) Magnesium chloride.

Dividing by the common factor (2) and applying crisis – cross.

Formula = CaCO3

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

(a) Quick lime (b) Hydrogen bromide (c) Baking powder (d) Potassium sulphate

Ans. (a) Quick lime is calcium oxide (CaO) Elements present: Calcium (Ca) and oxygen (O)

(b) Hydrogen bromide is HBr. Elements present: Hydrogen (H) and bromine (Br)

(c) Baking powder is sodium hydrogen carbonate (NaH CO3). Elements present: Sodium (Na), hydrogen (H), Carbon (C), and Oxygen (O).

(d) Potassium sulphate (K2SO4) Element present: Potassium (k), Sulphur (S) and Oxygen (O)

(e) Calcium carbonate is (CaCO3) Element present: Calcium (Ca), Carbon (C) and Oxygen (O)

Q. 6. Calculate the molar mass of the following substances :

(a) Ethyne

(b) Sulphur molecule, S8

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

(d) Hydrochloric acid, HCl (e) Nitric acid, HNO3

Ans. (a) Molar mass of ethyne, (C2H2) = 2 × atomic mass of C + 2 × atomic mass of

H = 2 × 12u + 2 × 1u = 26u

(b) Molar mass of sulphur, (S8) = 8 × atomic mass of S = 8 × 32u = 256u

(c) Molar mass of hydrochloric acid, (HCl) = 1 × atomic mass of

H + 1 × atomic mass of

Cl = 1 × 1u + 35.5 u = 36.5u

(d) Molar mass of nitric acid, (HNO3) = 1 × atomic mass of H + 1 × atomic mass of

N + 3 × atomic mass of O = 1 × 1u + 1 × 14u + 3 × 16u = 63u.

Q. 7. What is the mass of:

(a) 1 mole of nitrogen atom?

(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?

(c) 10 moles of sodium sulphite (Na2 SO3)? [Atomic mass N = 14u,

Al = 27u, Na = 23u, S = 32u, O = 16u]

Ans. (a) 1 mole of nitrogen atoms = 1 × gram atomic mass of nitrogen atom = 1 × 14g = 14g

(b) 4 moles of aluminium atoms = 4 × gram atomic mass of aluminium atoms = 4 × 27g = 108g

(c) 10 moles of sodium sulphite (Na2 CO3) = 10 (2 × gram atomic mass of Na + 1 × gram atomic mass of sulphur + 3 × (gram atomic mass of oxygen) = 10 (2 × 23g + 1 × 32g + 3 × 16g) = 10 (46g + 32g + 48g)

= 10 × 126g = 1260g.

Q. 8. Convert into mole :

(a) 12 g of oxygen gas

(b) 20 g of water

(c) 22 g of carbon dioxide

Ans. (a) 1 mole of oxygen (O2) = 2 × 16 g = 32 g  32 g oxygen = 1 mole

Q. 9. What is the mass of:

(a) 0.2 mole of oxygen atoms?

(b) 0.5 mole of water molecules?

Ans. (a) Mass of 0.2 mole atoms of oxygen = 0.2 mol × molar mass of oxygen atom = 0.2 mol × 16g/mol

= 3.2 g

(b) Mass of 0.5 mole molecules of water = 0.5 mol × molar mass of water = 0.5 mol × 18 g 1 mol = 9 g.

Q. 10. Calculate the number of aluminium ions present in 0.051 g of aluminium oxide. (At mass Al = 27u, O = 16u)

Ans. Molecular formula of aluminium oxide = Al2O3

Molecular mass of Al2O3 = 2 × 27 + 3 × 16 = 54 + 48 = 102g mol–1

= 5 × 10–4 mole One mole of Al2O3 contains 6.022 × 1023 formula unit of Al2O3  5 × 10–4 mole of Al2O3

\ will contain formula units of Al2O3 = (5 × 10–4) × (6.022 × 1023) = 30.11 × 1019 = 3.011 × 1020 Now,  one formula unit of Al2O3  contains 2 aluminium ions.

3.011 × 1020 formula units will contain aluminium ions = 2 × 3.011 × 1020

= 6.022 × 1020
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