Question

1. (a) State Graham’s law of diffusion
(b) Arrange the following gases He,CH4,and N2 in order of increasing rates of diffusion, give reasons for your answers. [H = 1,He = 4,C = 12,N = 14]
2. 60cm3 of hydrogen diffused through a porous membrane in 10 minutes. The same volume of a
gas G diffuses through the same membrane in 37.4 minutes. Determine the relative molecular
mass of G. [H = 1]

Answer 1

1. (a) According to Graham's law, the rate of diffusion or of effusion of a gas $r$ is inversely proportional to the square root of its molecular mass $M$:

$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$ .

(b) In order to arrange the gases in order of increasing rates of diffusion, we must calculate their molecular masses.

He: 4

CH₄: 12+1·4 = 16

N₂: 14·2 = 28.

The molecular masses of the gases increase in a row: He<CH₄<N₂. Therefore, the rate of diffusion increases in a row: N₂<CH₄<He.

2. The rate of the diffusion can be defines as an amount of gas passing through the area per unit of time: $r = V/t$. Making use of the analytical expression of Graham's law given in p.1 of the Question, the relative molecular mass of the gas G is:

$M_G = M_{H_2}·\frac{r_{H_2}^2}{r_G^2} = M_{H_2}·\frac{t_G^2}{t_{H_2}^2}$

$M_G = 1·2·\frac{37.4^2}{10^2} = 28$ g/mol.

With a high probability, the gas G is N₂, nitrogen.