Question

2.9 g of a gas at 95° C occupied the same volume as 0.184 g of dihydrogen at 17 °C, at the same pressure. What is the molar mass of the gas?

Answer 1

From the gas equation,
PV = (w /M)RT
where w = weight of gas.
T₁ = 273 + 95 = 368 K
and T₂ = 273 + 17 = 290 K
Substituting the given data in the gas equation, we get
PV = (2.9/M) × R × 368 ….. (i)
PV = (0.184/2) × R × 290 ….. (ii)
From these two equations, we can write
(2.9/M) × R × 368 = (0.184/2) × R × 290
By striking through R from both sides we get
(2.9/M) × 368 = (0.184/2) × 290
Or (2.9/M) = (0.092 × 290) / 368
Or M = 2.9 × 368/0.092 × 290
= 40 g/mol
Hence, the molar mass of the gas is 40 g mol^-1