The rate of diffusion of methane (CH_4) at a | vedclass

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(A) According to Graham's Law of diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$: $r \propt

Vedclass · Accepted Answer

$64$

Vedclass · Answer

(A) According to Graham's Law of diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.Given: $r_{CH_4} = 2 	imes r_X$.Using the ratio: $\frac{r_{CH_4}}{r_X} = \sqrt{\frac{M_X}{M_{CH_4}}}$.Substituting the values: $2 = \sqrt{\frac{M_X}{16}}$.Squaring both sides: $4 = \frac{M_X}{16}$.Therefore,$M_X = 4 	imes 16 = 64 \ g/mol$.

The rate of diffusion of methane $(CH_4)$ at a given temperature is twice that of gas $X$. The molecular weight of gas $X$ is:

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