At $27^{\circ} \mathrm{C}, 500 \mathrm{~mL}$ of helium diffuses in 30 minutes. What is the time (in hours)…

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?
  1. $240$
  2. $3$
  3. $2$
  4. $4$

Solution

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} $

Asked in: AP EAMCET 2004

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