An open beaker of water in equilibrium with water vapour is in a sealed container. When a few grams of glucose are added to the beaker of water, the rate at which water molecules:

At the start, pure water in the beaker is in dynamic equilibrium with the water vapour filling the sealed container. “Dynamic equilibrium” means that, every second, an equal number of molecules

$$\text{rate of evaporation} \;=\; \text{rate of condensation}$$

so the pressure of the vapour remains constant.

Now a few grams of glucose (a non-volatile solute) are dissolved in the water. Glucose cannot enter the vapour phase, therefore the solution obtained is no longer pure water. At this stage we recall Raoult’s law for an ideal solution of a non-volatile solute:

$$p_{\text{solution}} = x_{\text{water}}\; p^{\ast}_{\text{water}}$$

where

$$p_{\text{solution}}$$ is the vapour pressure of water over the solution,

$$x_{\text{water}} < 1$$ is the mole fraction of water (made smaller by the glucose),

$$p^{\ast}_{\text{water}}$$ is the vapour pressure of pure water at the same temperature.

Because $$x_{\text{water}}$$ has become less than 1, we have

$$p_{\text{solution}} < p^{\ast}_{\text{water}}.$$

Thus, immediately after adding glucose, the vapour already present in the sealed container is at the higher pressure $$p^{\ast}_{\text{water}}$$, while the new solution can support only the lower pressure $$p_{\text{solution}}$$. The system is therefore no longer in equilibrium: the pressure of the vapour is now larger than the equilibrium pressure corresponding to the solution.

Whenever the actual vapour pressure is higher than the equilibrium vapour pressure of the liquid present, the net process that restores equilibrium is condensation. Molecules leave the vapour phase and enter the liquid phase until the vapour pressure falls to $$p_{\text{solution}}$$. In other words, the rate at which water molecules leave the vapour (condense) temporarily increases.

Conversely, the rate at which molecules leave the solution (evaporate) momentarily decreases, because the escaping tendency of water from the solution has been lowered by the presence of glucose.

Therefore, the only correct statement is that the rate at which water molecules leave the vapour increases.

Hence, the correct answer is Option A.