Step 1: Understanding the Question:
Vapour pressure is a colligative property. Specifically, the relative lowering of vapour pressure is proportional to the concentration of solute particles.
Step 2: Key Formula or Approach:
Vapour pressure $(\text{V.P.}) \propto \frac{1}{\text{Lowering of V.P.}} \propto \frac{1}{\text{Effective Concentration } (i \times M)}$
where $i$ is the van't Hoff factor.
Step 3: Detailed Explanation:
Maximum vapour pressure means minimum lowering of vapour pressure, which occurs when there are the fewest number of particles in solution.
Assuming molarity $M = 0.1$ for all:
- (A) Glucose: Non-electrolyte, $i = 1$. Total particles $= 0.1 \times 1 = 0.1$.
- (B) $NaCl$: Dissociates into $Na^+ + Cl^-$, $i = 2$. Total particles $= 0.1 \times 2 = 0.2$.
- (C) $CaCl_2$: Dissociates into $Ca^{2+} + 2Cl^-$, $i = 3$. Total particles $= 0.1 \times 3 = 0.3$.
- (D) $AlCl_3$: Dissociates into $Al^{3+} + 3Cl^-$, $i = 4$. Total particles $= 0.1 \times 4 = 0.4$.
Glucose has the lowest effective concentration of particles, so it shows the maximum vapour pressure.
Step 4: Final Answer:
$0.1$ M Glucose solution shows maximum vapour pressure.