1. What Graham's law would predict

For two gases/vapours diffusing from opposite ends, Graham's law gives

If the first visible product forms where the two fronts meet, then the distances travelled satisfy

With $L=24\,\text{cm}$,

So Graham's law predicts the product should first appear at $16\,\text{cm}$ from the plug soaked in $X$.

But experimentally, $d$ is smaller than this. That means the product forms closer to the X end than Graham's law predicts. Hence, in reality, $X$ diffuses less effectively relative to $Y$ than Graham's law estimate suggests.

2. Why Graham's law is not exact here

Graham's law is most appropriate when comparing diffusion/effusion under simplified conditions. Here, both vapours diffuse through an inert gas already present at 1 atm. So each vapour undergoes repeated collisions mainly with inert-gas molecules.

Thus, the relative diffusion is controlled not only by molar masses of $X$ and $Y$, but also by how frequently each collides with the inert gas.

3. Compare molecular speeds of X and Y

At the same temperature,

Since

we have

So molecules of $X$ move faster and therefore collide more frequently with inert-gas molecules than molecules of $Y$ do.

4. Effect on diffusion distance

More frequent collisions with the inert gas hinder the net progress of $X$ more strongly. Therefore, compared to the Graham's law estimate, $X$ does not advance as far as expected. So the meeting point shifts towards the X end, making $d$ smaller.

This matches the observation.

5. Check options

• A: larger mean free path for $X$ than for $Y$
  This would help $X$ diffuse farther, making $d$ larger, not smaller. Incorrect.

• B: larger mean free path for $Y$ than for $X$
  While this could shift the point toward $X$, the more direct kinetic explanation here is collision frequency with inert gas.

• C: increased collision frequency of $Y$ with inert gas compared to $X$
  This would hinder $Y$ more, pushing meeting point away from $X$, so $d$ would increase. Incorrect.

• D: increased collision frequency of $X$ with inert gas compared to $Y$
  This hinders $X$ more, so the product appears closer to the $X$ end and $d$ becomes smaller. Correct.

6. Final answer

The correct option is