Benzene `(C_(6)H_(6), 78 g//mol)` and toluene `(C_(7)H_(8),92g//mol)` form an ideal solution. At `60^(@)C` the vapour pressure of pure benzene and pure toluene are `0.507` atm and `0.184` atm, respectively. The mole fraction of benezen in a solution of these two chemicals that has a vapour preesure of `0.350` atm at `60^(@)C`, will be :

Options

A. `0.514`
B. `0.690`
C. `0.486`
D. `0.190`

Answer

A

Explanation

`p_("solution")=p_(B)^(@)x_(B)+p_(T)^(@)x_(T)`
(T = toluene, B = Benzene)
`= p_(B)^(@)x_(B)+p_(T)^(@)(1-x_(B))`
`p_("solution")=p_(T)^(@)+(p_(B)^(@)-p_(T)^(@))x_(B)`
`0.350 =0.184 + (0.507 - 0.184)x_(B)`
`x_(B)=(0.166)/(0.323)=0.514`