Key principle: vapour pressure lowering

For a pure solvent, the vapour pressure is maximum.

If a non-volatile solute is dissolved in the same solvent, the vapour pressure of the solvent above the solution becomes lower.

By Raoult’s law: $p_{\text{solution}} = x_{\text{solvent}} \, p^{0}_{\text{solvent}}$ where $x_{\text{solvent}} < 1$, so $p_{\text{solution}} < p^{0}_{\text{solvent}}$

What happens in the sealed container?

The two open beakers are kept together in a closed container:

• one has pure solvent
• the other has solution of solvent + non-volatile solute

Since vapour pressure over pure solvent is higher, solvent molecules evaporate more readily from the pure solvent.

The vapour in the container will then condense preferentially into the solution, because the solution has lower equilibrium vapour pressure.

Net transfer of solvent

Therefore, there is a net transfer of solvent from the pure solvent beaker to the solution beaker.

So:

• volume of pure solvent decreases
• volume of solution increases

Check options

• A: The volume of the solution increases and the volume of the solvent decreases. ✅
• B: Solution does not change, solvent decreases. ❌
• C: Neither changes. ❌
• D: Solution decreases, solvent increases. ❌

Final answer

The correct option is: $\boxed{A}$