80 mL of O_2 takes 2 minute to pass through | vedclass

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(A) According to Graham's Law of Diffusion,the rate of diffusion $r$ is given by $r = \frac{V}{t} \propto \frac{1}{\sqrt{M}}$.For two gases,the ratio

Vedclass · Accepted Answer

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

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(A) According to Graham's Law of Diffusion,the rate of diffusion $r$ is given by $r = \frac{V}{t} \propto \frac{1}{\sqrt{M}}$.For two gases,the ratio is $\frac{r_1}{r_2} = \frac{V_1}{t_1} 	imes \frac{t_2}{V_2} = \sqrt{\frac{M_2}{M_1}}$.Given: $V_1 = 80 \ mL$,$t_1 = 2 \ min$,$M_1 (O_2) = 32 \ g/mol$,$t_2 = 3 \ min$,$M_2 (SO_2) = 64 \ g/mol$.Substituting the values: $\frac{80}{2} 	imes \frac{3}{V_2} = \sqrt{\frac{64}{32}}$.$40 	imes \frac{3}{V_2} = \sqrt{2}$.$V_2 = \frac{120}{\sqrt{2}} \ mL$.

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

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