Question:
20 mL of gas \(A\) and 10 mL of gas \(B\) diffuse through a porous membrane separately in 1 minute. If the vapor density of \(B\) is \(X\), what is the vapor density of \(A\)?
20 mL of gas \(A\) and 10 mL of gas \(B\) diffuse through a porous membrane separately in 1 minute. If the vapor density of \(B\) is \(X\), what is the vapor density of \(A\)?
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According to Graham's law, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass or vapor density:
\[
r\propto \frac{1}{\sqrt{VD}}.
\]
Updated On: Jun 26, 2026
- \(2X\)
- \(4X\)
- \(\dfrac{X}{4}\)
- \(\dfrac{X}{2}\)
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The Correct Option is C
Solution and Explanation
Step 1: Use Graham's law of diffusion.
According to Graham's law, \[ \frac{r_A}{r_B}=\sqrt{\frac{M_B}{M_A}} \] Since vapor density is proportional to molar mass, \[ \frac{r_A}{r_B}=\sqrt{\frac{VD_B}{VD_A}} \]
Step 2: Find the ratio of rates of diffusion.
Gas \(A\) diffuses \(20\text{ mL}\) in \(1\) minute, so \[ r_A=20 \] Gas \(B\) diffuses \(10\text{ mL}\) in \(1\) minute, so \[ r_B=10 \] Therefore, \[ \frac{r_A}{r_B}=\frac{20}{10}=2 \]
Step 3: Substitute vapor density values.
Given, \[ VD_B=X \] Let vapor density of gas \(A\) be \[ VD_A=y \] Using Graham's law, \[ 2=\sqrt{\frac{X}{y}} \]
Step 4: Square both sides.
\[ 4=\frac{X}{y} \] Therefore, \[ y=\frac{X}{4} \] So, \[ VD_A=\frac{X}{4} \]
Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{X}{4}} \]
According to Graham's law, \[ \frac{r_A}{r_B}=\sqrt{\frac{M_B}{M_A}} \] Since vapor density is proportional to molar mass, \[ \frac{r_A}{r_B}=\sqrt{\frac{VD_B}{VD_A}} \]
Step 2: Find the ratio of rates of diffusion.
Gas \(A\) diffuses \(20\text{ mL}\) in \(1\) minute, so \[ r_A=20 \] Gas \(B\) diffuses \(10\text{ mL}\) in \(1\) minute, so \[ r_B=10 \] Therefore, \[ \frac{r_A}{r_B}=\frac{20}{10}=2 \]
Step 3: Substitute vapor density values.
Given, \[ VD_B=X \] Let vapor density of gas \(A\) be \[ VD_A=y \] Using Graham's law, \[ 2=\sqrt{\frac{X}{y}} \]
Step 4: Square both sides.
\[ 4=\frac{X}{y} \] Therefore, \[ y=\frac{X}{4} \] So, \[ VD_A=\frac{X}{4} \]
Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{X}{4}} \]
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