Raoult's Law Example Problem

This example problem demonstrates how to use Raoult's Law to calculate the change in vapor pressure by adding a nonvolatile liquid to a solvent.

Problem:

What is the change in vapor pressure when 164 g of glycerin (C₃H₈O₃) is added to 338 mL of H₂O at 39.8 °C.
The vapor pressure of pure H₂O at 39.8 °C is 54.74 torr
The density of H₂O at 39.8 °C is 0.992 g/mL.

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

Step 1 Determine the mole fraction of solution

molar weight_glycerin (C₃H₈O₃) = 3(12)+8(1)+3(16) g/mol
molar weight_glycerin = 36+8+48 g/mol
molar weight_glycerin = 92 g/mol

moles_glycerin = 164 g x 1 mol/92 g
moles_glycerin = 1.78 mol

molar weight_water = 2(1)+16 g/mol
molar weight_water = 18 g/mol

density_water = mass_water/volume_water

mass_water = density_water x volume_water
mass_water = 0.992 g/mL x 338 mL
mass_water = 335.296 g

moles_water = 335.296 g x 1 mol/18 g
moles_water = 18.63 mol

Χ_solution = n_water/(n_water + n_glycerin)
Χ_solution = 18.63/(18.63 + 1.78)
Χ_solution = 18.63/20.36
Χ_solution = 0.91

Step 2 - Find the vapor pressure of the solution

P_solution = Χ_solventP⁰_solvent
P_solution = 0.91 x 54.74 torr
P_solution = 49.8 torr

Step 3 - Find the change in vapor pressure

Change in pressure is P_final - P_O
Change = 49.8 torr - 54.74 torr
change = -4.94 torr

Answer:

The vapor pressure of the water is reduced by 4.94 torr with the addition of the glycerin.