Estimation of Diffusion Distance Using Grahams Law

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 $M$ under identical conditions of temperature and pressure: $r \propto \frac{1}{\sqrt{M}}$

Since the vapours of $\mathbf{X}$ and $\mathbf{Y}$ start diffusing simultaneously from opposite ends of the tube of length $L = 24\text{ cm}$, the time $t$ taken by both gases to meet and form the product is the same.

Let:

• $d$ be the distance travelled by the vapour of $\mathbf{X}$.
• $L - d = 24 - d$ be the distance travelled by the vapour of $\mathbf{Y}$.

The ratio of their rates of diffusion is given by: $\frac{r_X}{r_Y} = \frac{d / t}{(24 - d) / t} = \frac{d}{24 - d}$

Applying Graham's law: $\frac{r_X}{r_Y} = \sqrt{\frac{M_Y}{M_X}}$

Given:

• $M_X = 10\text{ g mol}^{-1}$
• $M_Y = 40\text{ g mol}^{-1}$

Substitute the given values into the equation: $\frac{d}{24 - d} = \sqrt{\frac{40}{10}} = \sqrt{4} = 2$

Solving for $d$: $d = 2(24 - d)$ $d = 48 - 2d$ $3d = 48$ $d = 16\text{ cm}$

Thus, the estimated value of distance $d$ via Graham's law is $16\text{ cm}$, which corresponds to Option (C).