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Q.

Two volatile liquids ‘A’ and ‘B’ form an ideal solution. Vapour pressure of pure liquid ‘A’ is 800 mm Hg and pure liquid ‘B’ is 400 mm Hg. Total vapour pressure of A-B mixture is 720 mm Hg. Mole fraction of ‘A’ in vapour phase will be

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a

0.89

b

0.8

c

0.7

d

0.67

answer is A.

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Detailed Solution

$Applying Raoult' s law p_{total} = p_{A}^{0} X_{A} + p_{B}^{0} X_{B} 720 mm = 800 mm X_{A} + 400 mm (1 - X_{A}); X_{A} = 0.8, X_{B} = 0.2 Applying Dalton' s law in vapour phase p_{A} = p_{total} y_{A} (y_{A} = M . F . of' A' in vapour phase) y_{A} = \frac{800 (0.8)}{720} = 0.89;$

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