Question

20 mL of gas A and 10 mL of gas B diffuses through a porous membrane separately in 1 minute. If the vapor density of B is X, what is the vapor density of A?

• A. 2x
• B. 4x
• C. $\frac{x}{4}$
• D. $\frac{x}{2}$

Hint:

Graham's Law of Diffusion: $r \propto 1/\sqrt{M}$ or $r \propto 1/\sqrt{VD}$.
For two gases A and B: $\frac{r_A}{r_B} = \sqrt{\frac{M_B}{M_A}} = \sqrt{\frac{VD_B}{VD_A}}$.
Rate of diffusion $r = \text{Volume diffused / time taken}$.
Molar Mass $M = 2 \times \text{Vapor Density (VD)}$.

Answer 1

The Correct Option is C

According to Graham's Law of Diffusion, the rate of diffusion ($r$) of a gas is inversely proportional to the square root of its molar mass ($M$) or its vapor density ($VD$), assuming constant temperature and pressure. $r \propto \frac{1}{\sqrt{M}}$ or $r \propto \frac{1}{\sqrt{VD}}$ (since Molar Mass $M = 2 \times VD$). Rate of diffusion ($r$) can be expressed as Volume diffused ($V$) per unit time ($t$): $r = V/t$. Given: For gas A: Volume $V_A = 20$ mL, time $t_A = 1$ minute. Rate of diffusion of A: $r_A = \frac{V_A}{t_A} = \frac{20 \text{ mL}}{1 \text{ min}} = 20 \text{ mL/min}$. For gas B: Volume $V_B = 10$ mL, time $t_B = 1$ minute. Rate of diffusion of B: $r_B = \frac{V_B}{t_B} = \frac{10 \text{ mL}}{1 \text{ min}} = 10 \text{ mL/min}$. Let $VD_A$ be the vapor density of gas A and $VD_B$ be the vapor density of gas B. From Graham's Law: $\frac{r_A}{r_B} = \sqrt{\frac{VD_B}{VD_A}}$. We are given $VD_B = X$. We need to find $VD_A$. Substitute the rates of diffusion: $\frac{20 \text{ mL/min}}{10 \text{ mL/min}} = \sqrt{\frac{X}{VD_A}}$. $\frac{20}{10} = 2$. So, $2 = \sqrt{\frac{X}{VD_A}}$. Square both sides: $2^2 = \left(\sqrt{\frac{X}{VD_A}}\right)^2$. $4 = \frac{X}{VD_A}$. Now, solve for $VD_A$: $VD_A = \frac{X}{4}$. This matches option (c). \[ \boxed{\frac{X}{4}} \]