The molecular weight of a gas that diffuses t | vedclass

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Question

(A) 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 \pro

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(A) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molecular weight $M$: $r \propto \frac{1}{\sqrt{M}}$.Given: $M_1 = 64$,$r_2 = 2r_1$.Using the formula $\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$,we get:$\frac{r_1}{2r_1} = \sqrt{\frac{M_2}{64}}$$\frac{1}{2} = \sqrt{\frac{M_2}{64}}$Squaring both sides: $\frac{1}{4} = \frac{M_2}{64}$$M_2 = \frac{64}{4} = 16$.Thus,the molecular weight is $16$.

The molecular weight of a gas that diffuses twice as rapidly as the gas with molecular weight $64$ is:

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