1. Initial equilibrium

An open beaker of pure water is kept in a sealed container. At equilibrium:

• rate of evaporation of water from liquid to vapour
• rate of condensation of water from vapour to liquid

are equal.

So initially, $r_{\text{evap}} = r_{\text{cond}}$

2. Effect of adding glucose

Glucose is a non-volatile solute. When glucose is dissolved in water:

• mole fraction of water decreases
• escaping tendency of water molecules from the liquid surface decreases
• vapour pressure of water decreases

By Raoult’s law, $p_{\text{water}} = x_{\text{water}} p^\circ_{\text{water}}$ Since $x_{\text{water}}<1$, the vapour pressure becomes smaller than that of pure water.

3. Interpretation in terms of rates

The rate at which water molecules leave the solution corresponds to evaporation.

Because glucose lowers the escaping tendency of water, the evaporation rate decreases.

Thus, $r_{\text{leave solution}} \downarrow$

Immediately after adding glucose, the vapour above the solution is too much for the new lower equilibrium vapour pressure, so condensation becomes relatively greater until a new equilibrium is established.

4. Check options

• A: leaves the solution increases — False
• B: leaves the vapour increases — False
• C: leaves the vapour decreases — False in the intended equilibrium-shift sense
• D: leaves the solution decreases — True

5. Final answer

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