Question

A vessel of volume 100 mL
contains 10% O2 and 90% unknown gas. The gases diffuses in 86 s
through a small hole of the vessel. If pure oxygen under the same
condition diffuses in 75 s, find the molecular mass of the unknown
gas.​

Answer 1

Answer:

A vessel of 100 ml contains mixture of 10% of O2 & 90% of the unknown gas.

The mixture diffuses in = 86 s

Pure O₂ diffuses in = 75 s

The molecular mass of pure O₂ = 32 g/mol

Let the rate of diffusion of the mixture be “r1” and that of pure O₂ be “r2”. Also, let the molecular mass of O₂ be “M2”, the molecular mass of the mixture be “M1” and the molecular mass of the unknown gas be “x”  .

So, the molecular mass of the mixture,

M1 = [(90*x) + (32*10)] / 100 = 0.9x + 3.2

Using Graham’s Law of diffusion, we get

r1/r2 = √[M2/M1]

⇒ (1/86) / (1/75) = √[32/(0.9x + 3.2)]

⇒ (0.872)² =  [32/(0.9x + 3.2)]

⇒ 0.760 (0.9x + 3.2) = 32

⇒ 0.684x + 2.432 = 32

⇒ x = 29.568 / 0.684 = 43.22 g/mol

Thus, the molecular mass of the unknown gas is 43.22 g/mol.