80 , mL of O_2 takes 2 minutes to pass throug | vedclass

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(A) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $\frac{r_1}{r

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$\frac{120}{\sqrt{2}} \, mL$

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(A) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$.Given $M_{O_2} = 32 \, g/mol$ and $M_{SO_2} = 64 \, g/mol$.The rate $r = \frac{V}{t}$,so $\frac{V_{O_2}/t_{O_2}}{V_{SO_2}/t_{SO_2}} = \sqrt{\frac{M_{SO_2}}{M_{O_2}}}$.Substituting the values: $\frac{80/2}{V_{SO_2}/3} = \sqrt{\frac{64}{32}}$.$40 	imes \frac{3}{V_{SO_2}} = \sqrt{2}$.$V_{SO_2} = \frac{120}{\sqrt{2}} \, mL$.

$80 \, mL$ of $O_2$ takes $2$ minutes to pass through a hole. What volume of $SO_2$ will pass through the same hole in $3$ minutes?

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