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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
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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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Vapour pressure of pure A is 70 mm of Hg at 25oC.It forms an ideal solution with 'B' in which mole fraction of A is 0.8. If the vapour pressure of the solution is 84 mm of Hg at 25oC the vapour pressure of pure B at 25oC is
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