The molar volume of liquid benzene (density `0.877 g mL^(-1))` increases by a factor of `2750` as it vapourises 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 toluene at `20^(@)C` has a vapour pressure of 46.0 torr. Find the mole fraction of benzene in the vapour above the solution.

Answer

`chi_(benzene)^(V)=0.735`

Explanation

Number of moles `(n)` =`"Mass" (m)/"Molar mass" (Mw)`
Density =`"Mass" /"Volume"`
So, Volume =`"Mass" /"Density"`
=`"Number of moles" xx "Molar mass"/"Density"`
In vapour phase,
At `20^(@)C`, for 1 mol of benzene,
Volume=`(1xx78xx2750)/0.877=244583.80 mL =244.58 L`
Similarly, for 1 mol of toluene,
Volume=`(1xx92)/0.867 xx7720`
=`819192.61 mL =819.19 L`
As we know that, `p_(V)=nRT`
For benzene, `P_(B)^(@)=nRT/V=(1xx0.0821xx293)/(244.58)`atm brgt =0.098 atm
For toluene, `P_(T)^(@)=nRT/V=(1xx0.0821xx293)/(819.19)`atm
=0.029 atm
`P=P_(B)^(@).chi_(B)+P_(T)^(@).chi_(T)`
`:' chi_(B) +chi_(T)=1`
`:. chi_(T) =1-chi_(B)`
`P=P_(B)^(@).chi_(B) +P_(T)(1-chi_(B))`
Total vapour pressure =46 torr
=`4/760=0.060` atm
`0.060=0.098chi_(B)+0.029(1-chi_(B))`
`0.060=0.098chi_(B)+0.029-0.029chi_(B)`
`=0.060-0.029=0.098chi_(B)-0.029chi_(B)=0.069chi_(B)`
`:. chi_(B)=0.031/0.069=0.45`
(in liquid phase)
`chi_(B) + chi_(T)=1`
`:. chi_(T)=1-0.45=0.55`
(in liquid phase)
Mole fraction of benzene in vapour
Phase `chi'_(B)=p_(B)/p_(T)=(0.098xx0.45)/0.060`
=`0.735`.