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

## Raoult's Law Example

What is the expected vapor pressure when 58.9 g of hexane (C_{6}H_{14}) is mixed with 44.0 g of benzene (C_{6}H_{6}) 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 the vapor pressure equation:

P_{solution} = Χ_{solvent}P^{0}_{solvent}

where

P_{solution} is the vapor pressure of the solution

Χ_{solvent} is mole fraction of the solvent

P^{0}_{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 in order to be able to calculate the mole fraction of the components.

From the periodic table, the atomic masses of the carbon and hydrogen atoms in hexane and benzene are:

C = 12 g/mol

H = 1 g/mol

Use the molecular weights to find the number of moles of each component:

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. It doesn't matter which component you use to perform the calculation. In fact, a good way to check your work is to do the calculation for both hexane and benzene and then make sure they add up to 1.

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 by plugging the values into the equation:

P_{Total} = Χ_{hexane}P^{0}_{hexane} + Χ_{benzene}P^{0}_{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.