# physical chemistry

There are 4 questions and plzz tell reason for all question

1. The volume og 1.0g hydrogen at NTP is:

(1) 2.24 L (2) 22.4 L

(3) 1.12 L (4) 11.2 L

2. 11 grams of a gas occupy 5.6 L of volume at STP The gas is:

(1) NO (2) N2O4

(3) CO (4) CO2

3. At NTP , 5.6 L(liters) of gas weight 8g (grams). The vapour density of gas is:

(1) 32 (2) 40

(3) 16 (4) 8

4. The vapour densities of two gases are in ratio 1:3 . Their molecular masses are in ratio of:

(1) 1:3 (2) 1:2

(3) 2:3 (4) 3:1

1. 1g of hydrogen => mole of H2 = 1/2

Volume H2 at NTP = 1/2 x 22.4 L = 11.2 L

Ans (4)

2. At STP, 5.6 litres is occupied by 11g of a gas

Then, 22.4 litres is occupied by

(22.4 × 11)/5.6 = 44 g of that gas

So, the molecular weight of the gas = 44g

The molecular weight of CO2 (Carbon dioxide) is also 44g.

Ans (4)

3. Weight of 5.6 litre of gas = 8 gm

so weight of 22.4 litre = 8×22.4/ 5.6 = 32 gm

so molecular weight = 32 [ because weight of 22. 4 litre of any gas = molecular wt.]

we know vapour density = molecular wt/2 = 32/2 = 16

Ans (3)

4. We know that, molecular mass = 2 x vapour density

Hence, if vapour densities are in ratio 1:3 then their molecular masses will also be in the same ratio.

Ans (1)