The molecular weight of a gas which diffuses | vedclass

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Question

(B) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molecular weight $M$: $r_1/r_

Vedclass · Accepted Answer

$72$

Vedclass · Answer

(B) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molecular weight $M$: $r_1/r_2 = \sqrt{M_2/M_1}$.Given that the rate of the gas $r_g = \frac{1}{6} r_{H_2}$,we have $\frac{r_{H_2}}{r_g} = 6$.Using the formula $M_g = M_{H_2} 	imes (\frac{r_{H_2}}{r_g})^2$,where $M_{H_2} = 2 \ g/mol$.$M_g = 2 	imes (6)^2 = 2 	imes 36 = 72 \ g/mol$.

The molecular weight of a gas which diffuses through a porous plug at $1/6^{th}$ of the speed of hydrogen under identical conditions is

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