Question

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

Answer 1

Answer:

The volume of hydrogen that diffuses at the same time under condition is 199.04 ml.

Explanation:

Given data:

Volume of O₂, V1 = 50 ml

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

To find: Volume of hydrogen, V2

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

Answer 2

Answer:

200ml is ur answer... Dear