Freezing Point Depression Example Problem

Calculate the Freezing Point Depression Temperature

Frozen
Freezing Point Depression: Water will form ice at a lower temperature when a solute is added to the water. nikamata / Getty Images

This example problem demonstrates how to calculate freezing point depression. The example is for a solution of salt in water.

Quick Review of Freezing Point Depression

Freezing point depression is one of the colligative properties of matter, which means it is affected by the number of particles, not the chemical identity of the particles or their mass. When a solute is added to a solvent, its freezing point is lowered from the original value of the pure solvent.

It doesn't matter whether the solute is a liquid, gas, or solid. For example, freezing point depression occurs when either salt or alcohol are added to water. In fact, the solvent can be any phase, too. Freezing point depression also occurs in solid-solid mixtures.

Freezing point depression is calculated using Raoult's Law and the Clausius-Clapeyron Equation to write an equation called Blagden's Law. In an ideal solution, freezing point depression only depends on solute concentration.

Freezing Point Depression Problem

31.65 g of sodium chloride is added to 220.0 mL of water at 34 °C. How will this affect the freezing point of the water?
Assume the sodium chloride completely dissociates in the water.
Given: density of water at 35 °C = 0.994 g/mL
Kf water = 1.86 °C kg/mol

Solution:

To find the temperature change elevation of a solvent by a solute, use the freezing point depression equation:

ΔT = iKfm

where
ΔT = Change in temperature in °C
i = van 't Hoff factor
Kf = molal freezing point depression constant or cryoscopic constant in °C kg/mol
m = molality of the solute in mol solute/kg solvent.



Step 1 Calculate the molality of the NaCl

molality (m) of NaCl = moles of NaCl/kg water

From the periodic table, find the atomic masses of the elements:

atomic mass Na = 22.99
atomic mass Cl = 35.45
moles of NaCl = 31.65 g x 1 mol/(22.99 + 35.45)
moles of NaCl = 31.65 g x 1 mol/58.44 g
moles of NaCl = 0.542 mol

kg water = density x volume
kg water = 0.994 g/mL x 220 mL x 1 kg/1000 g
kg water = 0.219 kg

mNaCl = moles of NaCl/kg water
mNaCl = 0.542 mol/0.219 kg
mNaCl = 2.477 mol/kg

Step 2 Determine the van 't Hoff factor

The van 't Hoff factor, i, is a constant associated with the amount of dissociation of the solute in the solvent.

For substances which do not dissociate in water, such as sugar, i = 1. For solutes that completely dissociate into two ions, i = 2. For this example NaCl completely dissociates into the two ions, Na+ and Cl-. Therefore, i = 2 for this example.

Step 3 Find ΔT

ΔT = iKfm

ΔT = 2 x 1.86 °C kg/mol x 2.477 mol/kg
ΔT = 9.21 °C

Answer:

Adding 31.65 g of NaCl to 220.0 mL of water will lower the freezing point by 9.21 °C.