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Vapour pressure of pure water at 25°C is 23.8 mm Hg. The vapour pressure of a solution with 18 g glucose (M.W. = 180) in 90 g water:
A23.1 mm Hg
B23.3 mm Hg
C25 mm Hg
D22.4 mm Hg
Answer & Solution
Correct answer: B. 23.3 mm Hg
Moles glucose = 18/180 = 0.1. Moles water = 90/18 = 5. Mole fraction water = 5/5.1 ≈ 0.98. P = x_water × P_pure = 0.98 × 23.8 ≈ 23.3 mm Hg. (Glucose lowers vapour pressure of water.)
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