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

A strong electrolyte is one that dissociates completely in water, such as a strong acid or base or a salt.
A strong electrolyte is one that dissociates completely in water, such as a strong acid or base or a salt. Dave and Les Jacobs / Getty Images

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

Psolution = ΧsolventP0solvent where

Psolution is the vapor pressure of the solution
Χsolvent is mole fraction of the solvent
P0solvent 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 weightwater = 2(1)+16 g/mol
molar weightwater = 18 g/mol

densitywater = masswater/volumewater

masswater = densitywater x volumewater
masswater = 0.987 g/mL x 800 mL
masswater = 789.6 g

moleswater = 789.6 g x 1 mol/18 g
moleswater = 43.87 mol

Χsolution = nwater/(nwater + nsolute)
Χsolution = 43.87/(43.87 + 1.18)
Χsolution = 43.87/45.08
Χsolution = 0.97

Step 2 - Find the vapor pressure of the solution

Psolution = ΧsolventP0solvent
Psolution = 0.97 x 102.1 torr
Psolution = 99.0 torr

Step 3 - Find the change in vapor pressure

Change in pressure is Pfinal - PO
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.

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Helmenstine, Todd. "Raoult's Law Example Problem - Vapor Pressure and Strong Electrolyte." ThoughtCo, Mar. 23, 2017, thoughtco.com/raoults-law-and-strong-electrolyte-solution-609524. Helmenstine, Todd. (2017, March 23). Raoult's Law Example Problem - Vapor Pressure and Strong Electrolyte. Retrieved from https://www.thoughtco.com/raoults-law-and-strong-electrolyte-solution-609524 Helmenstine, Todd. "Raoult's Law Example Problem - Vapor Pressure and Strong Electrolyte." ThoughtCo. https://www.thoughtco.com/raoults-law-and-strong-electrolyte-solution-609524 (accessed December 11, 2017).