50 ml of oxygen diffuse under certain conditions through a porous membrane. The volume of hydrogen that diffuses in the same time under the same conditions is

Volume of O₂, V₁ = 50 ml

Time taken by oxygen and hydrogen to diffuse is same i.e., t₁ = t₂

Let the rate of diffusion of O₂ & H₂ be “r₁” & “r₂” respectively.  

The molar mass of O₂, M₁ = 31.988 g/mol

The molar mass of H₂, M₂ = 2.02 g/mol

Also, the rate of diffusion of gas, r = V/t ….(i)

Using Graham’s Law,

r1/r2 = √[M2/M1] ……. (ii)

from (i) & (ii), we get

[V₁/t₁] / [V₂/t₂] = √[2.02/31.988]

Or, 50 / V₂ = 0.2512

Or, V₂ = 50 / 0/2512 = 199.04 ml