At `40^(@)C` the vapour pressure of pure liquids, benzene and toluene, are `160 mm Hg` and `60 mm Hg` respectively. At the same temperature, the vapour pressure of an equimolar solution of the liquids, assuming the ideal solution will be:

Options

A. 140 mm of Hg
B. 110 mm of Hg
C. 220 mm of Hg
D. 100 mm of Hg

Answer

B

Explanation

`P_(s)=P_(B)^(@)+X_(T)(P_(P)^(@)-P_(P)^(@))=160+(1)/(2)(60-160)=160-50=110`.