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 (C_{3}H_{8}O_{3}) is added to 338 mL of H_{2}O at 39.8 °C.

The vapor pressure of pure H_{2}O at 39.8 °C is 54.74 torr

The density of H_{2}O 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_{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**Step 1** Determine the mole fraction of solution

molar weight_{glycerin} (C_{3}H_{8}O_{3}) = 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_{solvent}P^{0}_{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.