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

166 torr

Step-by-step 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 * \text{304 torr} = \text{ 152 torr}\\p_(2) = 0.5 * \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 $\boxed{\textbf{166 torr}}$}`