A mixture of chlorobenzene and wate (immiscible) boils at $90.3^oC$ at an external pressure of 740.2 mm. The vapour pressure of pure water at $90.3^oC$ is 530.1 mm. Calculate the % composition of distillate :
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$H_2O=$ 35%
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$H_2O=$ 22%
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$H_2O=$ 29%
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$H_2O=$ 71%
C
Correct answer
Explanation
Dalton's law for steam distillation: P_total = P_water + P_chlorobenzene. 740.2 = 530.1 + P_chlorobenzene => P_chlorobenzene = 210.1. Mass ratio = (P1 * M1) / (P2 * M2). Molar masses: H2O=18, Chlorobenzene=112.5. Mass ratio = (530.1 * 18) / (210.1 * 112.5) = 9541.8 / 23636.25 = 0.403. % H2O = 0.403 / 1.403 = 28.7%.