Which of the following statements about the composition of the vapour over an ideal `1:1` mol mixture of benzene and toluene is correct? Assume that the temperature is constant at `25^(@)C`. (Given: vapour pressure Date at `25^(@)C`, benzene=12.8 kP, toluene=3.85 kPa)

Options

A. Not enough information is given to make predicition.
B. The vapour will contain a higher precentage of benzene.
C. The vapour will contain a higher percentage of toluene
D. The vapour will contain equal amounts of benzene and toluene.

Answer

2

Explanation

Note that benzene is more volatile than toluene. According to Konowaloff's rule, at only fixed temperature, vapour phase above a solution is always richer in the more volatile component as compared to solution phase.
In other words, mole fraction of the more volatile component is always greater in the vapour phase than in the solution phase.
Alternatively, vapour phase is relatively richer in the component whose addition to the solution phase results in an increase in the total vapour pressure.
Since we are given `1:1` mol mixture of benzene `(A)` and toluene `(T)`, we have
`X_(A)=1/2` and `X_(B)=1/2`
According to Raoult's law
`P_(total)=P_(A)^(0)X_(A)+P_(B)^(0)X_(B)`
`=(12.8 kpa)(1//2)+(3.85 kpa)(1//2)`
`=(6.4 kpa)+(1.93 kpa)`
`=8.33 kpa`
If `y_(A)` and `y_(B)` are the mole fractions of the benzene and toluenes respectivaly in the vapour phase, then
`y_(A)=P_(A)/P_("total")=(P_(A)^(0)X_(A))/P_("total")(6.4 kpa)/(8.33 kpa)`
`=0.77`
and `y_(b) =1 - 0.77=0.23`
Thus, at equilibrium, vapour phase will be rich in benzene, the more volatile component.