Question

At a certain temperature, ethane has a vapor pressure of 304 torr and propane has a vapor pressure of 27 torr. What is the mole fraction of ethane in the vapor above a solution containing equal molar amounts of ethane and propane?

Answer 1

Answer:

166 torr

Explanation:

Let’s call ethane Component 1 and propane Component 2.

According to Raoult’s Law,  

$p_{1} = \chi_{1}p_{1}^{\circ}\\p_{2} = \chi_{2}p_{2}^{\circ}$

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.

Data:

p₁° = 304 torr

p₂° =   27 torr

n₁ = n₂

1. Calculate the mole fraction of each component

χ₁ = n₁/(n₁ + n₂)

χ₁ = n₁/n₁ + n₁)

χ₁ = n₁/(2n₁)

χ₁ = ½

χ₁ = 0.0.5

χ₂ = 1- χ₁ = 1- 0.5 = 0.5

2. Calculate the vapour pressure of the mixture

$p_{1} = 0.5 \times \text{304 torr} = \text{ 152 torr}\\p_{2} = 0.5 \times \text{27 torr} = \text{ 13.5 torr}\\p_{\text{tot}} = p_{1} + p_{2} = \text{152 torr + 13.5 torr} = \textbf{166 torr}$

$\text{The vapour pressure above the solution is \boxed{\textbf{166 torr}} }$