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

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 percent

B

34 mol percent

C

48 mol percent

D

50 mol percent

Text Solution

Verified by Experts

`P_(T)=P_(A)^(@)X_(A)+P_(B)^(@)X_(B)implies760=520x_(A)+P_(B)^(@)(1-X_(A))" "impliesX_(A)=0.5`
Thus mole % of `A=50%`
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