Q.57 A binary mixture of benzene and toluene under vapour-liquid equilibrium at
80 oC follows ideal Raoult’s law. At this condition, the saturation pressures of
benzene and toluene are 101 kPa and 40 kPa, respectively. If the mole fraction
of benzene in the liquid phase is 0.6, the corresponding mole fraction of benzene
in the vapour phase will be _______.
(Round off to two decimal places)
Using Raoult’s law, with benzene liquid mole fraction of 0.6 and saturation
pressures of 101 kPa (benzene) and 40 kPa (toluene),
the vapor mole fraction of benzene is found to be 0.79.
Raoult’s Law Basics
Raoult’s law states that the partial pressure of a component in an ideal mixture equals the
product of its liquid mole fraction and its pure-component saturation pressure:
Pi = xi Pi∘
The total pressure is the sum of all partial pressures, and the vapor mole fraction is given by:
yi = Pi / P
Benzene–toluene behaves nearly ideally, making Raoult’s law applicable.
Step-by-Step Calculation
- Benzene liquid mole fraction, xB = 0.6
- Toluene liquid mole fraction, xT = 0.4
Benzene partial pressure:
0.6 × 101 = 60.6 kPa
Toluene partial pressure:
0.4 × 40 = 16 kPa
Total pressure:
60.6 + 16 = 76.6 kPa
Vapor mole fraction of benzene:
yB = 60.6 / 76.6 = 0.79
Why 0.79 Is Correct
Benzene has a higher saturation pressure than toluene, making it more volatile.
As a result, benzene is enriched in the vapor phase:
yB > xB
Incorrect answers often arise from ignoring volatility differences or using the wrong total
pressure. The correct value rounded to two decimal places is 0.79.
Common Errors Explained
| Error Type | Calculation | Result | Why Incorrect |
|---|---|---|---|
| Direct equality | yB = xB | 0.60 | Ignores volatility difference |
| Wrong component | yT = xT | 0.40 | Calculates toluene fraction instead |
| Wrong total pressure | 60.6 / 80.6 | 0.75 | Incorrect pressure summation |
| Correct | 60.6 / 76.6 | 0.79 | Correct use of Raoult’s law |
