Question

Which of the following aqueous solutions will show maximum vapour pressure at 300K?

• A. \(1\) M NaCl
• B. \(1\) M \(\text{CaCl}_2\)
• C. \(1\) M \(\text{AlCl}_3\)
• D. \(1\) M \(\text{C}_{12}\text{H}_{22}\text{O}_{11}\)

Hint:

Vapour pressure is highest for the solution with the fewest dissolved particles.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Raoult's law says the vapour pressure of a solution is \(p = x_{\text{solvent}}\,p^{\circ}\). Adding a non-volatile solute lowers the mole fraction of the solvent, so the vapour pressure falls. This is a colligative property: it depends on the number of solute particles, not on what they are.

Step 2: Key Formula or Approach:
More particles means a bigger fall in vapour pressure. So the solution with the fewest particles shows the highest vapour pressure. The number of particles from one formula unit is given by the van't Hoff factor \(i\).

Step 3: Detailed Explanation:
All solutions are \(1\) M, so we compare \(i\).
\(\text{NaCl} \to \text{Na}^+ + \text{Cl}^-\), so \(i = 2\).
\(\text{CaCl}_2 \to \text{Ca}^{2+} + 2\text{Cl}^-\), so \(i = 3\).
\(\text{AlCl}_3 \to \text{Al}^{3+} + 3\text{Cl}^-\), so \(i = 4\).
Sucrose, \(\text{C}_{12}\text{H}_{22}\text{O}_{11}\), is a molecular solid that does not ionise in water, so \(i = 1\).
Effective particle concentrations are about \(2\) M, \(3\) M, \(4\) M and \(1\) M.

Step 4: Why the other options are wrong:
The solution of \(\text{AlCl}_3\) has the most particles and the greatest lowering, so it has the lowest vapour pressure. \(\text{CaCl}_2\) and \(\text{NaCl}\) also remove more solvent from the surface than sucrose does. So (A), (B) and (C) all show a smaller vapour pressure than (D).

Final Answer:
The \(1\) M sucrose solution has the fewest particles, so it has the maximum vapour pressure. This is option (D). \[ \boxed{\text{Option (D)}} \]