Raoult's Law Example Problem

This example problem demonstrates how to use Raoult's Law to calculate the of vapor pressure of two volatile solutions mixed together.

Problem:

What is the expected vapor pressure when 58.9 g of hexane (C₆H₁₄) is mixed with 44.0 g of benzene (C₆H₆) at 60.0 °C?

Given:
Vapor pressure of pure hexane at 60 °C is 573 torr.
Vapor pressure of pure benzene at 60 °C is 391 torr.

Solution

Raoult's Law can be used to express the vapor pressure relationships of solutions containing both volatile and nonvolatile solvents.
Raoult's Law is expressed by

P_solution = Χ_solventP⁰_solvent

where

P_solution is the vapor pressure of the solution
Χ_solvent is mole fraction of the solvent
P⁰_solvent is the vapor pressure of the pure solvent

When two or more volatile solutions are mixed, each pressure component of the mixed solution is added together to find the total vapor pressure.

P_Total = P_{solution A} + P_{solution B} + ...

Step 1 Determine the number of moles of each solution

From the periodic table:
C = 12 g/mol
H = 1 g/mol

molar weight of hexane = 6(12) + 14(1) g/mol
molar weight of hexane = 72 + 14 g/mol
molar weight of hexane = 86 g/mol

n_hexane = 58.9 g x 1 mol/86 g
n_hexane = 0.685 mol

molar weight of benzene = 6(12) + 6(1) g/mol
molar weight of benzene = 72 + 6 g/mol
molar weight of benzene = 78 g/mol

n_benzene = 44.0 g x 1 mol/78 g
n_benzene = 0.564 mol

Step 2 - Find mole fraction of each solution

Χ_hexane = n_hexane/(n_hexane + n_benzene)
Χ_hexane = 0.685/(0.685 + 0.564)
Χ_hexane = 0.685/1.249
Χ_hexane = 0.548

Since there are only two solutions present and the total mole fraction is equal to one:

Χ_benzene = 1 - Χ_hexane
Χ_benzene = 1 - 0.548
Χ_benzene = 0.452

Step 3 - Find the total vapor pressure

P_Total = Χ_hexaneP⁰_hexane + Χ_benzeneP⁰_benzene
P_Total = 0.548 x 573 torr + 0.452 x 391 torr
P_Total = 314 + 177 torr
P_Total = 491 torr

Answer:

The vapor pressure of this solution of hexane and benzene at 60 °C is 491 torr.