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At 80oC, the vapour pressure of pure liquid 'A' is 520 mmHg and that of pure liquid ‘B’ is 1000 mmHg. If a mixture solution of 'A' and ' B' boils at 80oC and 1 atm pressure, the amount of 'A' in the mixture is: (1 atm = 760 mmHg):

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a

52 mol percent

b

34 mol percent

c

48 mol percent

d

50 mol percent

answer is D.

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Detailed Solution

Pm=PAoXA+PBoXB760=520XA+1000(1-XA)760 = 520XA+1000-1000XA1000-760 = 1000XA-520XA240 = 480XAXA=240480∴XA=0.5=50%

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At 80oC, the vapour pressure of pure liquid 'A' is 520 mmHg and that of pure liquid ‘B’ is 1000 mmHg. If a mixture solution of 'A' and ' B' boils at 80oC and 1 atm pressure, the amount of 'A' in the mixture is: (1 atm = 760 mmHg):