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

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$20 \ s$ : $O_2$

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(B) According to Graham's law of effusion,the rate of effusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \frac{1}{\sqrt{M}}$.Since $r = \frac{V}{t}$,for the same volume $V$,we have $\frac{V}{t} \propto \frac{1}{\sqrt{M}}$,which implies $t \propto \sqrt{M}$.Therefore,the ratio of times is $\frac{t_1}{t_2} = \sqrt{\frac{M_1}{M_2}}$.Given $t_{H_2} = 5 \ s$ and $M_{H_2} = 2 \ g/mol$:For $He$ $(M = 4)$: $t = 5 \sqrt{\frac{4}{2}} = 5\sqrt{2} \ s \approx 7.07 \ s$.For $O_2$ $(M = 32)$: $t = 5 \sqrt{\frac{32}{2}} = 5 	imes 4 = 20 \ s$.For $CO$ $(M = 28)$: $t = 5 \sqrt{\frac{28}{2}} = 5\sqrt{14} \ s \approx 18.7 \ s$.For $CO_2$ $(M = 44)$: $t = 5 \sqrt{\frac{44}{2}} = 5\sqrt{22} \ s \approx 23.45 \ s$.Thus,the correct match is $20 \ s$ for $O_2$.

$X \ mL$ of $H_2$ gas effuses through a hole in a container in $5 \ s$. The time taken for the effusion of the same volume of the gas specified below under identical conditions is:

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