Practice question
Question
The vapor pressure of pure water is 30 mm Hg at a certain temperature. A solution with a non-volatile solute has a vapor pressure of 28.5 mm Hg. If the solute’s molality is 0.5 mol/kg, what is the mass of water in the solution?
Explanation
(p⁰ - p/p⁰) = xsolute . (30 - 28.5/30) = 0.05 = (nsolute/nsolute + nwater) . Molality = (nsolute/wwater) = 0.5 . Assume wwater = 1 kg , then nsolute = 0.5 . nwater = (1000/18) ≈ 55.56 . xsolute = (0.5/0.5 + 55.56) ≈ 0.0089 , adjust wwater = (0.5 × 18/0.05) = 180 g .
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