Question

The vapour pressure of a solution containing 2 moles of a solute in 2 moles of water (vapour pressure of pure water = 24mmHg) is

• A. 24 mm Hg
• B. 32 mm Hg
• C. 48 mm Hg
• D. 12 mm Hg

Answer 1

Answer: (D)

12 mmHg

Answer 2

Solution:

1. Total moles in solution:

   Moles of solute = 2
   Moles of water = 2
   Total moles = 2 + 2 = 4

2. Mole fraction of water (solvent):

   $X_{\text{water}} = \frac{2}{4} = 0.5$

3. Vapour pressure of the solution (using Raoult's Law):

   $P_{\text{solution}} = X_{\text{water}} \times P^0 = 0.5 \times 24\, \text{mmHg} = 12\, \text{mmHg}$