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$20 \mathrm{~mL}$ of gas $A$ and $10 \mathrm{~mL}$ of gas $B$ diffuses through a porous membrane separately in 1 minute. If the vapour density of $B$ is $X$, what is the vapour density of $A$ ?
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The correct answer is:
$\frac{x}{4}$
Rate of diffusion $\propto \frac{1}{\sqrt{\text { Molecular mass }}}$
Molecular mass $=2 \times$ vapour density
Rate of diffusion $\propto \frac{1}{\sqrt{2 \times \text { vapour density }}}$
$$
\begin{aligned}
& \frac{r_A}{r_B}=\frac{V_A / t_A}{V_B / t_B}=\frac{\sqrt{2 \times \text { vapour density of } B}}{\sqrt{2 \times \text { vapour density of } A}} \\
& \frac{r_A}{r_B}=\frac{20 / 1}{10 / 1}=\frac{\sqrt{2 \times x}}{\sqrt{2 \times \text { vapour density of } A}}
\end{aligned}
$$
Vapour density of $A=x / 4$
Molecular mass $=2 \times$ vapour density
Rate of diffusion $\propto \frac{1}{\sqrt{2 \times \text { vapour density }}}$
$$
\begin{aligned}
& \frac{r_A}{r_B}=\frac{V_A / t_A}{V_B / t_B}=\frac{\sqrt{2 \times \text { vapour density of } B}}{\sqrt{2 \times \text { vapour density of } A}} \\
& \frac{r_A}{r_B}=\frac{20 / 1}{10 / 1}=\frac{\sqrt{2 \times x}}{\sqrt{2 \times \text { vapour density of } A}}
\end{aligned}
$$
Vapour density of $A=x / 4$
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