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If the vapour pressure of water at a particular temperature is 25 mm Hg, and 1 mol of a non-volatile solute is dissolved in 9 mol of water, the lowering in vapour pressure is:
A0.25 mm Hg
B2.5 mm Hg
C5.0 mm Hg
D22.5 mm Hg
Answer & Solution
Correct answer: B. 2.5 mm Hg
$x_2 = 1/(1+9) = 0.10$. $\Delta P = x_2 P_1^0 = 0.10 \times 25 = 2.5$ mm Hg.
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