Raoult's Law Example Problem - Vapor Pressure and Strong Electrolyte

This example problem demonstrates how to use Raoult's Law to calculate the change in vapor pressure by adding a strong electrolyte to a solvent. Raoult's Law relates the vapor pressure of a solution on the mole fraction of the solute added to a chemical solution.

Vapor Pressure Problem

What is the change in vapor pressure when 52.9 g of CuCl₂ is added to 800 mL of H₂O at 52.0 °C.
The vapor pressure of pure H ₂O at 52.0 °C is 102.1 torr
The density of H₂O at 52.0 °C is 0.987 g/mL.

Solution Using Raoult's Law

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
CuCl₂ is a strong electrolyte. It will completely dissociate into ions in water by the reaction:
CuCl₂(s) → Cu²⁺(aq) + 2 Cl^-
This means we will have 3 moles of solute added for every mole of CuCl₂ added.
From the periodic table:
Cu = 63.55 g/mol
Cl = 35.45 g/mol
molar weight of CuCl₂ = 63.55 + 2(35.45) g/mol
molar weight of CuCl₂ = 63.55 + 70.9 g/mol
molar weight of CuCl₂ = 134.45 g/mol
moles of CuCl₂ = 52.9 g x 1 mol/134.45 g
moles of CuCl₂ = 0.39 mol
Total moles of solute = 3 x (0.39 mol)
Total moles of solute = 1.18 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 = 0.987 g/mL x 800 mL
mass_water = 789.6 g
moles_water = 789.6 g x 1 mol/18 g
moles_water = 43.87 mol
Χ_solution = n_water/(n_water + n_solute)
Χ_solution = 43.87/(43.87 + 1.18)
Χ_solution = 43.87/45.08
Χ_solution = 0.97

Step 2

Find the vapor pressure of the solution
P_solution = Χ_solventP⁰_solvent
P_solution = 0.97 x 102.1 torr
P_solution = 99.0 torr

Step 3

Find the change in vapor pressure
Change in pressure is P_final - P_O
Change = 99.0 torr - 102.1 torr
change = -3.1 torr

Answer

The vapor pressure of the water is reduced by 3.1 torr with the addition of the CuCl₂.