2.5 g of a non-volatile, non-electrolyte is dissolved in 100 g of water at 25° C. The solution showed a boiling point elevation by 2° C. Assuming the solute concentration is negligible with respect to the solvent concentration, the vapor pressure of the resulting aqueous solution is _________ mm of Hg (nearest integer)
[Given : Molal boiling point elevation constant of water (K_b)=0.52 K . kg mol⁻¹, 1 atm pressure =760 mm of Hg, molar mass of water =18 g mol⁻¹]

Solution

Chemistry ma’am · 4 steps

Get the molality from the boiling point rise, then turn it into a mole fraction for Raoult's law.

1. Molality first. From Δ T_b = K_b m, the molality is m = 2/0.52 = 3.846 mol kg⁻¹. You do not need the 2.5 g of solute at all.
2. Pure water's pressure. The elevation is measured against the boiling point of pure water, where water's vapour pressure equals 1 atm. So take p° = 760 mm Hg.

   (p° - p)/p° = x₂ ≈ n₂/n₁ = m × M₁/1000

3. Put in numbers. Since the solute is dilute, x₂ ≈ 3.846 × 18/1000 = 0.0692. The lowering is 760 × 0.0692 = 52.6 mm Hg.
4. So. p = 760 - 52.6 = 707.4 mm Hg, which rounds to 707.

The common slip: 53. Stopping at the lowering of vapour pressure gives 53, not 707. Subtract it from 760 to get the solution's vapour pressure.

The idea to keep. Colligative properties all link through molality. For a dilute solution, x₂ ≈ m M₁/1000, so Δ T_b gives you the relative lowering of vapour pressure directly.

Answer from the NTA final answer key. Solution by PrepWiser.