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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

`52 mol%`

B

`34 mol%`

C

`48mol%`

D

`50 mol%`

Text Solution

Verified by Experts

The correct Answer is:
D
At boiling point `P_(sys) = P_(sorr')`
`760 = 520 xx x_(A) + 1000 (1-x_(A))`
`x_(A) = (240)/(480) = 0.50` or `50%`
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