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(D) According to Graham's Law of Diffusion,the rate of diffusion $r$ is given by $r = \frac{V}{t} = \frac{k}{\sqrt{M}}$,where $V$ is volume,$t$ is tim

Vedclass · Accepted Answer

$45.26 \ mL$ of $He$

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(D) According to Graham's Law of Diffusion,the rate of diffusion $r$ is given by $r = \frac{V}{t} = \frac{k}{\sqrt{M}}$,where $V$ is volume,$t$ is time,and $M$ is molar mass.Since the time $t$ is the same for all gases,the volume $V$ diffused is directly proportional to $\frac{1}{\sqrt{M}}$.Therefore,$\frac{V_1}{\sqrt{M_1}} = \frac{V_2}{\sqrt{M_2}}$.For $H_2$: $V_1 = 64 \ mL$,$M_1 = 2 \ g/mol$.$\frac{64}{\sqrt{2}} = \frac{V_2}{\sqrt{M_2}} \implies V_2 = \frac{64 	imes \sqrt{M_2}}{\sqrt{2}} = 64 	imes \sqrt{\frac{M_2}{2}}$.Checking option $(D)$ for $He$ $(M_2 = 4)$:$V_2 = 64 	imes \sqrt{\frac{4}{2}} = 64 	imes \sqrt{2} \approx 64 	imes 1.414 = 90.5 \ mL$ (Incorrect).Checking option $(C)$ for $D_2$ $(M_2 = 4)$:$V_2 = 64 	imes \sqrt{\frac{4}{2}} = 90.5 \ mL$ (Incorrect).Checking option $(B)$ for $CH_4$ $(M_2 = 16)$:$V_2 = 64 	imes \sqrt{\frac{16}{2}} = 64 	imes \sqrt{8} = 64 	imes 2.828 = 181 \ mL$ (Incorrect).Re-evaluating the calculation: The rate $r = \frac{V}{t} \propto \frac{1}{\sqrt{M}}$. For same $t$,$V \propto \frac{1}{\sqrt{M}}$.$V_{gas} = V_{H_2} 	imes \sqrt{\frac{M_{H_2}}{M_{gas}}} = 64 	imes \sqrt{\frac{2}{M_{gas}}}$.For $He$ $(M=4)$: $V = 64 	imes \sqrt{0.5} = 64 	imes 0.707 = 45.26 \ mL$. Thus,option $(D)$ is correct.

Under identical conditions,the time required for the diffusion of $64 \ mL$ of $H_2$ gas is equal to the time required for the diffusion of which of the following?

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