Two volatile liquids A and B are mixed in 1 : 1 mole ratio. Vapour pressure of pure liquids A
& B are 200 mm of Hg and 600 mm of Hg respectively. Total vapour pressure of the mixture was found to be 350 mm of Hg. The correct statement about the liquid mixture is

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Shows + ve deviation from Raoults law

Mixture can be separated by fractional distillation

It is an ideal solution

Shows - ve deviation from Raoults law

answer is C.

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Detailed Solution

$P_{S} = P_{A}^{0} X_{A} + P_{B}^{0} X_{B}$

$= 200 \times \frac{1}{2} + 600 \times \frac{1}{2} = 100 + 300 = 400$

Solution vapour pressure 350 mm of Hg is less than ideal solution vapour pressure hence it shows negative deviation from ideal solution and forms maximum boiling point azerotrope.

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