Question

A mixture of 60 mol % n-propylcyclohexane and 40 mol % n-propylbenzene is distilled through a simple distillation apparatus; assume that no fractionation occurs during the distillation. the boiling temperature is found to be 157 degrees celsius (760 torr) as the first small amount of distillate is collected. the standard vapor pressures of n-propylcyclohexane and n-propyl-benzene are known to be 769 torr and 725 torr, respectively, at 1567.3 degrees celsius. calculate the percentage of each of the two components in the first few drops of distillate.

Answer 1

Answer:

61 mole % propylcyclohexane and 39 mole % propylbenzene

Explanation:

For convenience, let’s call propylcyclohexane Component 1 and propylbenzene Component 2.

According to Raoult’s Law,  

$p_{1} = \chi_{1}p_{1}^{\circ}$ and

$p_{2} = \chi_{2}p_{2}^{\circ}$

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.

So,  

$p_{1} = 0.60 \times \text{769 torr} = \text{ 461 torr}$

$p_{2} = 0.40 \times \text{725 torr} = \text{ 290 torr}$

$p_{\text{tot}}$ = p₁ + p₂= 461 torr + 290 torr = 751 torr

∴ In the vapour

$\chi_{1} = \frac{p_{1}}{p_{\text{tot}} } = \frac{\text{461 torr} }{\text{751 torr}} = 0.61$

χ₂ = 1 – χ₁ = 1 - 0.61 = 0.39