Question

9. Which of the following prove law of multiple proportion?
a) H,SO,,H,S b) CCl4, CH4 c) H2O, HO
d) N2O, NO​

Answer 1

4. How many atoms are present in a

(i) H2S molecule and

(ii) PO43– ion?

Ans. (i) H2S → 3 atoms are present

(ii) PO43– → 5 atoms are present

NCERT TEXTBOOK PAGE 40

1. Calculate the molecular masses of H2, O2, C12, CO2, CH4, C2H6, C2H4, NH3, CH3OH.

Ans. The molecular masses are:

H2 ⇒ 1 × 2 → 2u

O2⇒ 16 × 2 → 32u

Cl2 ⇒ 35.5 × 2 → 71 u

CO2 ⇒ 1 × 12 + 2 × 16 = 12 +32 = 44 u

CH4 ⇒ 1 × 12 + 4 × 1 = 16 u

C2H6 ⇒�n2 × 12 + 6 × 1 = 30 u

C2H4 ⇒ (2 × 12) + (4 × 1) = 28 u

NH3 ⇒ (1 × 14) + (3 × 1) = 17 u

CH3OH ⇒ 12 + (3 × 1) + 16 + 1 = 32 u

2. Calculate the formula unit masses of ZnO, Na2O, K2Co3 given atomic masses of Zn=65 u, Na = 23 u, K = 39 u, C = 12u, and O = 16 u.

Ans. The formula unit mass of

(i) ZnO = 65 u + 16 u =.81 u

(ii) Na2O = (23 u × 2) + .16 u : = 46u+ 161.u.

= 62 u

(iii) K2CO3 = (39 u × 2) + 12 u + 16 u × 3

= 75 u + 12u = 48u = 138u

NCERT TEXTBOOK PAGE 42

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

Ans. 1 mole of carbon atoms 6.022 × 1023 atoms = 12 g

Mass of 1 atom = ?

∴ Mass of 1 atom of carbon

=1.99 × 10–23 g

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. 23 g of Na = 6.022 × 1023 atoms (1 mole).

100 g of Na = ?

= 26.182 ×1023 = 2.6182×1024 atoms

56 g of Fe = 6.022 × 1023 atoms

100 g o Fe = ?

100 g of Na contain → 2.618 × 1024 atoms

100 g of Fe contain → 1.075 × 1024 atoms

∴ 100 g of Na contains more atoms.

QUESTIONS FROM NCERT TEXTBOOK

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. Boron and oxygen compound → Boron + Oxygen

0.24 g → 0:096g + 0.144g

Percentage composition of the compound

For boron: 0.24g → 0.096 g

100 g → ?

For oxygen:

0.24 g → 0.144 g of oxygen

100 g → ?

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 formed when. 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemicai combination will govern your answer?

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

It shows that 12 g of carbon burns 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 �V 8 = 42 g of 02. The answer governs the law of constant proportion.

3. What are polyatomic ions? Give eaamples:

Ans. 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–, SO2–4, CO32–.

4. Write the chemical formulae of the following:

(a) Magnesium chloride (b) Caldum oxide

(c) Copper nitrate (d) Aluminium chloride

(e) Calcium carbonate.

Ans. (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 NO3

Change → +2 –1

Formula → Cu(NO3)2

(d) Aluminium chloride

Symbol → Al Cl

Change → +3 –1

Formula → A1Cl3

(d) Calcium carbonate

Symbol → Ca CO3

Change → +2 –2

Formula → CaCO3