Question

What is the vapour pressure of a solution containing 0.1 mol of non volatile solute dissolved in 16.2 g water? ( $P_{t} ^ 0 = 24 \, \text{mmHg}$. molar mass of water 18 g/mol)

• A. 12.4 mmHg
• B. 15.7 mmHg
• C. 18.1 mmHg
• D. 21.6 mmHg

Answer 1

Answer: (D)

21.6 mmHg

Answer 2

1. Calculate moles of water:

   Moles of water = 16.2 g / 18 g/mol = 0.9 mol

2. Determine total moles in the solution:

   Total moles = Moles of water + Moles of solute = 0.9 + 0.1 = 1 mol

3. Apply Raoult’s Law:

   The mole fraction of the solute = 0.1/1 = 0.1

   Relative lowering of vapour pressure = Mole fraction of solute = (P° - P) / P°

   So, P = P° (1 - 0.1) = 0.9 × P°

   Given P° = 24 mmHg,

   P = 0.9 × 24 = 21.6 mmHg