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At 80^(@)C the vapour pressure of pure l...

At `80^(@)C` the vapour pressure of pure liquid 'A' is 520 mm Hg and that of pure liquid 'B' is 1000 mm Hg. If a mixture solution of 'A' and 'B' boils at `80^(@)C` and 1 atm pressure, the amount of 'A' in the mixture is (1 atm `= 760 mm Hg)`

A

34 mol percent

B

48 mol percent

C

50 mol percent

D

52 mol percent

Text Solution

Verified by Experts

The correct Answer is:
C
`P_A^@=520 P_B^@=1000 `
`P_T=P_A^@x_A+P_B^@x_B`
`760+520x_A +1000(1-x_A)`
`760+520x_A +1000-1000x_A`
`480x_A=240,x_A=1/2`
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