X mL of H_2 gas effuse through a hole in a c | vedclass

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(B) According to Graham's Law of effusion,the rate of effusion $r \propto \frac{1}{\sqrt{M}}$.Since $r = \frac{V}{t}$,for the same volume $V$,we have

Vedclass · Accepted Answer

$20 \ seconds : O_2$

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(B) According to Graham's Law of effusion,the rate of effusion $r \propto \frac{1}{\sqrt{M}}$.Since $r = \frac{V}{t}$,for the same volume $V$,we have $\frac{V}{t_1} 	imes \sqrt{M_1} = \frac{V}{t_2} 	imes \sqrt{M_2}$,which simplifies to $\frac{t_2}{t_1} = \sqrt{\frac{M_2}{M_1}}$.Given $t_1 = 5 \ s$ for $H_2$ $(M_1 = 2 \ g/mol)$,we calculate $t_2$ for each gas:For $O_2$ $(M_2 = 32 \ g/mol)$: $t_2 = 5 	imes \sqrt{\frac{32}{2}} = 5 	imes \sqrt{16} = 5 	imes 4 = 20 \ s$.Thus,the correct option is $20 \ seconds : O_2$.

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

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