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At $27^{\circ} \mathrm{C}, 500 \mathrm{~mL}$ of helium diffuses in 30 minutes. What is the time (in hours) taken for $1000 \mathrm{~mL}$ of $\mathrm{SO}_2$ to diffuse under same experimental conditions?
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Verified Answer
The correct answer is:
$4$
Given, $r_{\mathrm{He}}=\frac{500}{30} \mathrm{~mL} / \mathrm{min}$
$$
r_{\mathrm{SO}_2}=\frac{1000}{t} \mathrm{~mL} / \mathrm{min}
$$
$$
\begin{gathered}
M_{\mathrm{He}}=4 \\
M_{\mathrm{SO}_2}=64
\end{gathered}
$$
From Graham's law
$$
\begin{aligned}
\frac{r_{\mathrm{He}}}{r_{\mathrm{SO}_2}} & =\sqrt{\frac{M_{\mathrm{SO}_2}}{M_{\mathrm{He}}}} \\
\frac{500}{30} \times \frac{t}{1000} & =\sqrt{\frac{64}{4}} \\
\Rightarrow \quad \frac{t}{60} & =4 \\
t & =240 \mathrm{~min}=4 \mathrm{hrs} .
\end{aligned}
$$
$$
r_{\mathrm{SO}_2}=\frac{1000}{t} \mathrm{~mL} / \mathrm{min}
$$
$$
\begin{gathered}
M_{\mathrm{He}}=4 \\
M_{\mathrm{SO}_2}=64
\end{gathered}
$$
From Graham's law
$$
\begin{aligned}
\frac{r_{\mathrm{He}}}{r_{\mathrm{SO}_2}} & =\sqrt{\frac{M_{\mathrm{SO}_2}}{M_{\mathrm{He}}}} \\
\frac{500}{30} \times \frac{t}{1000} & =\sqrt{\frac{64}{4}} \\
\Rightarrow \quad \frac{t}{60} & =4 \\
t & =240 \mathrm{~min}=4 \mathrm{hrs} .
\end{aligned}
$$
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