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For a dilute solution containing 2.5 g of a non–volatile non–electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2 °C. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take K b = 0.76 K kg mol -1) ?

Jayant Kumar , 12 Years ago
Grade 11
anser 2 Answers
Saurabh Koranglekar

Vapour Pressure of a Dilute Solution

Given:

  • Mass of solute = 2.5 g (non-volatile, non-electrolyte)
  • Mass of water = 100 g = 0.1 kg
  • Boiling point elevation, \( \Delta T_b = 2^\circ C \)
  • Boiling point elevation constant, \( K_b = 0.76 \, \text{K kg mol}^{-1} \)
  • Vapour pressure of pure water at 100°C, \( P_0 = 760 \, \text{mm Hg} \)

Step 1: Calculate Molality

m = ΔT_b / K_b = 2 / 0.76 ≈ 2.63 mol/kg

Step 2: Calculate Moles of Solute

Moles of solute = molality × kg of solvent = 2.63 × 0.1 = 0.263 mol

Step 3: Calculate Moles of Solvent (Water)

Moles of water = 100 g / 18 g/mol ≈ 5.56 mol

Step 4: Mole Fraction of Solute

x_solute = 0.263 / (0.263 + 5.56) ≈ 0.0452

Step 5: Apply Raoult’s Law

P = P₀ × (1 − x_solute) = 760 × (1 − 0.0452) ≈ 760 × 0.9548 ≈ 725.65 mm Hg
✅ Vapour Pressure of the Solution ≈ 726 mm Hg
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