Answer:
(a) Benzene = 0.26; toluene = 0.74
(b) Benzene = 0.55
Explanation:
1. Calculate the composition of the solution
For convenience, let’s call benzene Component 1 and toluene Component 2.
According to Raoult’s Law,
[tex]p_{1} = \chi_{1}p_{1}^{\circ}\\p_{2} = \chi_{2}p_{2}^{\circ}[/tex]
where
p₁ and p₂ are the vapour pressures of the components above the solution
χ₁ and χ₂ are the mole fractions of the components
p₁° and p₂° are the vapour pressures of the pure components.
Note that
χ₁ + χ₂ = 1
So,
[tex]\begin{array}{rcl}p_{\text{tot}} &=& p_{1} + p_{2}\\p_{\text{tot}} & = & \chi_{1}p_{1}^{\circ} + (1 - \chi_{1})p_{2}^{\circ}\\36 & = & \chi_{1}\times 75 + (1 - \chi_{1}) \times 22 \\36 & = & 75\chi_{1} + 22 -22\chi_{1}\\14 & = & 53\chi_{1}\\\chi_{1} & = & \mathbf{0.26}\\\end{array}[/tex]
χ₁ = 0.26 and χ₂ = 0.74
2. Calculate the mole fraction of benzene in the vapour
In the liquid,
p₁ = χ₁p₁° = 0.26 × 75 mm = 20 mm
∴ In the vapour
[tex]\chi_{1} = \dfrac{p_{1}}{p_{\text{tot}} } = \dfrac{\text{20 mm} }{\text{36 mm}} = 0.55[/tex]
Note that the vapour composition diagram below has toluene along the horizontal axis. The purple line is the vapour pressure curve for the vapour. Since χ₂ has dropped to 0.45, χ₁ has increased to 0.55.
