Question

A 100 ml mixture contains 72% by volume metane ans rest an unkniwn gas x was kept in vessel. due to crack the mixture effused out.21l was lost and the remaining contained 68.35% metahne by volume. what ia the milecular mass of gas z

Answer 1

100/72%
now the answer will be *68.35%/21

Answer 2

Required formula, ratio of rate of diffusion;

v₁ / v₂ = √m₂ / √m₁

Initially, volume of methane(gas 1) = 72 ml

Thus, volume of gas x = (100-72) = 28 ml

After 21 ml is lost, Volume of methane = (100-21) X 68.35% = 54 ml

Thus the volume of gas x = (100-21) - 54 = 25 ml

Loss of methane = (72-54) = 18 ml

Loss of gas x = (28-25) = 3 ml

Molecular mass of methane = 16 g/mol

Therefore, putting the values in the formula;

18/3 = √m₂/√16

or, 6 = √m₂/4

or, √m₂ = 24

or, m₂ = 576

Molecular mass of the gas x is 576.