**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.