Question

A gaseous mixture of O2 and x containing 20% of x, diffused through a small hole in 134 sec while pure O2 takes 124 secs to diffuse through the same hole. Find molecular weight of x.

Answer 1

Let the molecular mass of x be m.

20% of this is :

0.2m

Molecular weight of oxygen gas = 32

Let the molecular weight of the mixture be n

The total molecular weight of the mixture is :

n

By Graham's law of diffusion :

134/124 = √(32)/(n)

1.081² = 32/(n)

1.1678 = 32/(n)

1.1678(n) = 32

1.1678n  = 32

n = 32/1.1678

n = 27.40

20/100 × 27.40 = 5.48

The molecular mass of x = 5.48g