300 mL of a gas 'x' of molar mass 32 g mol | vedclass

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

Vedclass · Accepted Answer

$424$

Vedclass · Answer

(D) According to Graham's Law of diffusion,the rate of effusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.Since the time taken is the same for both gases,the ratio of the rates of effusion is equal to the ratio of the volumes effused $(V)$: $\frac{r_x}{r_{CH_4}} = \frac{V_x}{V_{CH_4}}$.Therefore,$\frac{V_x}{V_{CH_4}} = \sqrt{\frac{M_{CH_4}}{M_x}}$.Given: $V_x = 300 \ mL$,$M_x = 32 \ g \ mol^{-1}$,$M_{CH_4} = 16 \ g \ mol^{-1}$.Substituting the values: $\frac{300}{V_{CH_4}} = \sqrt{\frac{16}{32}} = \sqrt{\frac{1}{2}} = \frac{1}{1.414}$.$V_{CH_4} = 300 	imes 1.414 = 424.26 \ mL \approx 424 \ mL$.

$300 \ mL$ of a gas '$x$' of molar mass $32 \ g \ mol^{-1}$ is effused in $25 \ s$. What volume of methane would effuse in the same time (in $mL$)?

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