Question

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:

• A. The volume of the solution and the solvent does not change
• B. The volume of the solution increases and the volume of the solvent decreases
• C. The volume of the solution decreases and the volume of the solvent increases
• D. The volume of the solution does not change and the volume of the solvent decreases

Answer 1

The correct option is B The volume of the solution increases and the volume of the solvent decreases

Consider beaker I contains the solvent and beaker 2 contains the solution. Let the vapour pressure of the beaker I be $P^{\circ }$and the vapour pressure of beaker II be $P^s$ . According to Raoult’s law, the vapour pressure of the solvent ($P^{\circ }$) is greater than the vapour pressure of the solution $\left(P^s\right)$

$(P^{\circ }>P^s)$

Due to a higher vapour pressure, the solvent flows into the solution. So volume of beaker II would increase.

In a closed beaker, both the liquids on attaining equilibrium with the vapour phase will end up having the same vapour pressure. Beaker II attains equilibrium at a lower vapour pressure and so in its case, condensation will occur so as to negate the increased vapour pressure from beaker I, which results in an increase in its volume.

On the contrary, since particles are condensing from the vapour phase in beaker II, the vapour pressure will decrease. Since beaker I at equilibrium attains a higher vapour pressure evaporation will be favoured more so as to compensate for the decreased vapour pressure, as mentioned in the previous statement.