The molar volume of liquid bezene (density `= 0.877 g mL^(-1)`) increase by a factor of `2750` as it vaporises at `20^(@)C` and that of liquid toluene (density `0.867 g mL^(-1)`) increases by a factor of `7720` at `20^(@)C`. A solution of benzene and tuluene at `20^(@)C` has a vapour pressure of `46.0` torr. Find the mole fraction of benzene in the vapour above the solution.

Explanation

In vapour phase,
`"1 mole of benzene, i.e., 78 g has volume at "20^(@)C=(78)/(0.877)xx"2750 mL"`
`"1 mole of toluene, i.e., 92 g has volume at "20^(@)C=(92)/(0.867)xx"7720 mL"`
`"Applying , PV = n RT"`
`"For benzene vapour, "(P^(@))/(760)"(atm)"xx(78xx2750)/(0.877xx1000)(L)=1xx"0.0821 L atm K"^(-1)"mol"^(-1)xx293 K`
`"or "P_(B)^(@)=74.74mm`
`"For toluene vapour, "(P^(@))/(760)xx(92xx7720)/(0.867xx1000)=1xx0.0821xx293`
`"or "P_(T)^(@)=22.37mm`
`" "P_("total")=x_(B)xxP_(B)^(@)+x_(T)xxP_(T)^(@)=x_(B)P_(B)^(@)+(1-x_(B))xxP_(T)^(@)`
`therefore" "46=x_(B)=(74.74)+(1-x_(B))(22.37)`
This on solving gives `x_(B)=0.45` (in the liquid phase)
`therefore" "x_(T)=1-0.45=0.55` (in the liquid phase)
`"In vapour phase, "P_(B)=x_(B)P_(B)^(@)=0.45xx74.74=36.63mm`
`" "P_(T)=x_(T)P_(T)^(@)=0.55xx22.37=12.30 mm`
`"Mole fraction of benzene in vapour phase "(y_(B))=(P_(B))/(P_(B)+P_(T))(36.63)/(36.63+12.30)=0.75`