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Two volatile liquids ‘A’ and ‘B’ are mixed in the mole ratio 2 : 3. Vapour pressure of pure liquid ‘A’ is 600 mm Hg and pure liquid ‘B’ is 400 mm Hg. Mole fraction of ‘B’ in vapour phase will be
detailed solution
Correct option is A
GivennA:nB=2 : 3 XA=25; XB=35; Applying Raoult's law pA=pA0XA; pB=pB0XB pA=60025=240 mmHg pB=40035=240 mmHg ptotal=pA+pB=480 mmHg Applying Dalton's law in vapour phase pB=ptotalyByB=M.F. of 'B' in vapour phase yB=240480=0.5Talk to our academic expert!
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