At `25^@C`the vapour prssure of benzene ,`C_(6)H_(6)(78g//mol e)`,is 93.2 torr and that of toluene ,`C_(7)H_(8)(92g//mol)`,is `28.2`torr .A solution of `1.0`mole fo `C_(6)H_(6)`and `1.0` mol of `C_(7)H_(8)`id prepared. Calculate the mole fraction of `C_(6)H_(6)`in the vapour above this solution (assume the solution is ideal.)

Options

A. `0.607`
B. `0.768`
C. `0.232`
D. `0.393`

Answer

B

Explanation

`Y_(B)=(P_(B)^(@)xxX_(B))/(P)`
`Y_(B)=(93.2xx0.5)/(P)`
`Y_(T)=(28.2xx0.5)/(P)`
`:.(Y_(B))/(Y_(T))=3.3&Y_(B)+Y_(T)=1`
on solving `Y_(B)=0.768`