Q.

Two volatile liquids ‘A’ and ‘B’ are mixed in the mole ratio 1 : 4. Vapour pressure of pure liquid ‘A’ is 800 mmHg and pure liquid ‘B’ is 100 mmHg. Mole fraction of ‘B’ in vapour phase will be

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0.2

0.8

0.33

0.67

answer is C.

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Detailed Solution

GivennA:nB=1 : 4 XA=15; XB=45; Applying Raoult's law pA=pA0XA;pB=pB0XB pA=80015=160 mm Hg pB=10045=80 mm Hg ptotal=pA+pB=240 mm Hg Applying Dalton's law in vapour phase pB=ptotalyB yB=M.F. of 'B' in vapour phase yB=80240=0.33 ;

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