Question:

The ratio of rate of diffusion of SO$_2$ to CH$_4$ is

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Remember: \( r \propto \frac{1}{\sqrt{M}} \)
Use molar masses in g/mol in Graham’s law.
Updated On: May 19, 2025
  • 1:1
  • 1:2
  • 2:1
  • 4:1
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The Correct Option is B

Solution and Explanation

From Graham’s Law: \[ \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \quad \text{(Rate inversely proportional to molar mass)} \] Molar masses: SO$_2$ = 64, CH$_4$ = 16 \[ \Rightarrow \frac{r_{\text{CH}_4}}{r_{\text{SO}_2}} = \sqrt{\frac{64}{16}} = \sqrt{4} = 2 \Rightarrow \frac{r_{\text{SO}_2}}{r_{\text{CH}_4}} = \frac{1}{2} \]
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