ICSE Class 10 Answered

Given:

P= 1140 mm Hg

Density = D = 2.4 g / L

T = 273 ⁰C = 273+273 = 546 K

M = ?

We know that, at STP, the volume of one mole of any gas is 22.4 L

Hence we have to find out the volume of the unknown gas at STP.

First apply Charle’s law.

We have to find out the volume of one liter of unknown gas at standard temperature 273 K

V₁= 1 L T₁ = 546 K

V₂=? T₂ = 273 K

V₁/T₁ = V₂/ T₂

V₂ = (V₁ x T₂)/T₁

= (1 L x 273 K)/546 K

= 0.5 L

We have found out the volume at standard temperature. Now we have to find out the volume at standard pressure.

Apply Boyle’s law.

P ₁ = 1140 mm Hg V₁ = 0.5 L

P₂ = 760 mm Hg V₂ = ?

P₁ x V₁ = P₂ x V₂

V₂ = (P₁ x V₁)/P₂

= (1140 mm Hg x 0.5 L)/760 mm Hg

= 0.75 L

Now, 22.4 L is the volume of 1 mole of any gas at STP, then 0.75 L is the volume of X moles at STP

X moles = 0.75 L / 22.4 L

= 0.0335 moles

The original mass is 2.4 g

n = m / M

0.0335 moles = 2.4 g / M

M = 2.4 g / 0.0335 moles

M= 71.6 g / mole

Hence, the gram molecular mass of the unknown gas is 71.6 g.