Vapour pressure of pure water is 100 (arbitrary units). Add 1 mole non-volatile solute to 9 moles water. New vapour pressure is:
A91
B10
C100
D90
Answer & Solution
Correct answer: D. 90
x_water = 9/10 = 0.9. By Raoult's law: P = 0.9 × 100 = 90 units. Lowering by 10 units (vapour pressure lowering).
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