50 mL of hydrogen diffuses out through a sma | 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 $M$: $r \propto \f

Vedclass · Accepted Answer

$64$

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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 $M$: $r \propto \frac{1}{\sqrt{M}}$.For two gases,$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$.Given $r_{H_2} = \frac{50 \ mL}{20 \ min} = 2.5 \ mL/min$ and $M_{H_2} = 2 \ g/mol$.For oxygen,$M_{O_2} = 32 \ g/mol$ and volume $V = 40 \ mL$.Let the time taken be $t$. Then $r_{O_2} = \frac{40}{t}$.Substituting the values: $\frac{2.5}{40/t} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.$\frac{2.5 	imes t}{40} = 4 \implies 2.5 	imes t = 160$.$t = \frac{160}{2.5} = 64 \ min$.

$50 \ mL$ of hydrogen diffuses out through a small hole from a vessel in $20 \ min$. The time needed for $40 \ mL$ of oxygen to diffuse out is ............. $min$.

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