Q.

Two volatile liquids ‘A’ and ‘B’ are mixed in the mole ratio 4 : 1. Vapour pressure of pure liquid ‘A’ is 400 mm Hg and pure liquid ‘B’ is 800 mm Hg. Mole fraction of ‘B’ in vapour phase will be

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0.63

0.45

0.67

0.33

answer is D.

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Detailed Solution

GivennA:nB=4 : 1 XA=45; XB=15; Applying Raoult's law pA=pA0XA; pB=pB0XB pA=40045=320 mmHg pB=80015=160 mmHg ptotal=pA+pB=480 mmHg Applying Dalton's law in vapour phase pB=ptotalyByB=M.F. of 'B' in vapour phase yB=160480=0.33

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