Question:
300 mL of a gas \(X\) of molar mass \(32\ \text{g mol}^{-1}\) is effused in 25 seconds. What volume of methane would effuse in the same time?
300 mL of a gas \(X\) of molar mass \(32\ \text{g mol}^{-1}\) is effused in 25 seconds. What volume of methane would effuse in the same time?
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According to Graham's law, the rate of effusion is inversely proportional to the square root of molar mass:
\[
r\propto \frac{1}{\sqrt{M}}
\]
Lighter gases effuse faster than heavier gases.
Updated On: Jun 18, 2026
- 624 mL
- 524 mL
- 32 mL
- 424 mL
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Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Apply Graham's law of effusion.
According to Graham's law, \[ \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \] Since the time is the same, the volume effused is proportional to the rate of effusion. Thus, \[ \frac{V_{CH_4}}{V_X} = \sqrt{\frac{M_X}{M_{CH_4}}} \]
Step 2: Substitute the given values.
Given, \[ M_X=32\ \text{g mol}^{-1} \] and for methane, \[ M_{CH_4}=16\ \text{g mol}^{-1} \] Therefore, \[ \frac{V_{CH_4}}{300} = \sqrt{\frac{32}{16}} \] \[ = \sqrt{2} \] \[ V_{CH_4} = 300\times 1.414 \] \[ = 424.2\ \text{mL} \]
Step 3: Final conclusion.
Hence, \[ \boxed{V_{CH_4}\approx 424\ \text{mL}} \] Therefore, the correct option is (4).
According to Graham's law, \[ \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \] Since the time is the same, the volume effused is proportional to the rate of effusion. Thus, \[ \frac{V_{CH_4}}{V_X} = \sqrt{\frac{M_X}{M_{CH_4}}} \]
Step 2: Substitute the given values.
Given, \[ M_X=32\ \text{g mol}^{-1} \] and for methane, \[ M_{CH_4}=16\ \text{g mol}^{-1} \] Therefore, \[ \frac{V_{CH_4}}{300} = \sqrt{\frac{32}{16}} \] \[ = \sqrt{2} \] \[ V_{CH_4} = 300\times 1.414 \] \[ = 424.2\ \text{mL} \]
Step 3: Final conclusion.
Hence, \[ \boxed{V_{CH_4}\approx 424\ \text{mL}} \] Therefore, the correct option is (4).
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