If X , mL of H_2 gas diffuses through a hole | vedclass

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

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$20 \, s : O_2$

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(B) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$ of the gas: $r \propto \frac{1}{\sqrt{M}}$.Since $r = \frac{V}{t}$,for the same volume $V$,$t \propto \sqrt{M}$.Therefore,$\frac{t_1}{t_2} = \sqrt{\frac{M_1}{M_2}}$.Given $t_1 = 5 \, s$ for $H_2$ $(M_1 = 2 \, g/mol)$.For option $B$: $M_2 = 32 \, g/mol$ $(O_2)$.$t_2 = t_1 	imes \sqrt{\frac{M_2}{M_1}} = 5 	imes \sqrt{\frac{32}{2}} = 5 	imes \sqrt{16} = 5 	imes 4 = 20 \, s$.

If $X \, mL$ of $H_2$ gas diffuses through a hole in a container in $5 \, s$,what is the time taken for the diffusion of the same volume of another gas under identical conditions?

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