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)`

Options

A. 48 mol percent
B. 50 mol percent
C. 52 mol percent
D. 34 mol percent

Answer

B

Explanation

`P_("total")=(at 80^(@)C)=760mm`
`P_("total")=x_(A)p_(A)^(@)+x_(B)p_(B)^(@)=x_(A)p_(A)^(@)+(1-x_(A))p_(B)^(@)`
`=p_(B)^(@)+x_(A)(p_(A)^(@)-p_(B)^(@))`
`therefore" "1000+x_(A)(520-1000)=760`
`"or "480x_(A)=240`
`"or "x_(A)=0.50,` i.e., 50 mol percent.