Two open beakers one containing a solvent and the other containing a mixture of that solvent with a non volatile solute are together sealed in a container. Over time:

We have two separate open beakers inside one closed container. The first beaker contains only the pure solvent, while the second beaker contains a solution made from the same solvent and a non-volatile solute.

A non-volatile solute, by definition, has practically zero vapour pressure. Therefore the vapour present above the solution is contributed only by the solvent molecules present in that solution.

According to Raoult’s law for an ideal liquid solution, the partial vapour pressure of the solvent in the solution $$P_{\text{solution}}$$ is given by

$$$P_{\text{solution}} = X_{\text{solvent}}\;P^{0}_{\text{solvent}}$$$

where

$$$X_{\text{solvent}} = \text{mole fraction of the solvent in the solution}$$$
$$$P^{0}_{\text{solvent}} = \text{vapour pressure of the pure solvent at the same temperature.}$$$

Because a part of the liquid in the solution has been replaced by a non-volatile solute, we clearly have $$X_{\text{solvent}} < 1.$$ Hence

$$$P_{\text{solution}} = X_{\text{solvent}}\;P^{0}_{\text{solvent}} < P^{0}_{\text{solvent}}.$$$

So, above the beaker containing the pure solvent the vapour pressure equals $$P^{0}_{\text{solvent}},$$ while above the solution the vapour pressure is the smaller value $$P_{\text{solution}}.$$

Both beakers are in the same closed container, so the space above them is common. Vapour always flows spontaneously from the region of higher pressure to the region of lower pressure until equilibrium is reached. Therefore solvent vapour will migrate from the region above the pure solvent to the region above the solution.

Now, whenever extra vapour condenses into the solution beaker, those solvent molecules turn back into liquid and join the solution. At the same time, because vapour is leaving the surface of the pure solvent beaker faster than it is re-entering, the liquid level in that beaker must fall (net evaporation).

Step by step, the visible consequences are:

1. The pure solvent beaker loses liquid volume — it keeps evaporating.

2. The solution beaker gains liquid volume — condensed solvent adds to it.

This process continues until the mole fraction $$X_{\text{solvent}}$$ in the solution becomes high enough that its vapour pressure equals the vapour pressure of the diminishing pure solvent beaker, or until the pure solvent is exhausted.

So, the volume of the solution increases while the volume of the pure solvent decreases.

Hence, the correct answer is Option A.