Freezing Point Depression Example Problem

This example problem demonstrates how to calculate freezing point depression using 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 thefreezing point of the water?
Assume thesodium chloride completely dissociates in the water.
Given: density of water at 35 °C = 0.994 g/mL
K_f water = 1.86 °C kg/mol

Solution

To find thetemperature change elevation of a solvent by a solute, use the freezing point depression equation:
ΔT = iK_fm
where
ΔT = Change in temperature in °C
i = van ‘t Hoff factor
K_f = 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
m_NaCl = moles of NaCl/kg water
m_NaCl = 0.542 mol/0.219 kg
m_NaCl = 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 intotwo 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 = iK_fm
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.

Limitations of Freezing Point Depression Calculations

Calculating freezing point depression has practical applications, such as making ice cream and drugs and de-icing roads. However, the equations are only valid in certain situations.

• The solute must be present in much lower quantities than the solvent. Freezing point depression calculations apply to dilute solutions.
• The solute must be non-volatile. The reason is that freezing point occurs when the vapor pressure of the liquid and solid solvent are at equilibrium.

Sources

• Atkins, Peter (2006). Atkins’ Physical Chemistry. Oxford University Press. pp. 150–153. ISBN 0198700725.
• Aylward, Gordon; Findlay, Tristan (2002). SI Chemical Data (5th ed.). Sweden: John Wiley & Sons. p. 202. ISBN 0-470-80044-5.
• Ge, Xinlei; Wang, Xidong (2009). “Estimation of Freezing Point Depression, Boiling Point Elevation, and Vaporization Enthalpies of Electrolyte Solutions”. Industrial & Engineering Chemistry Research. 48 (10): 5123. doi:10.1021/ie900434h
• Mellor, Joseph William (1912). “Blagden’s Law”. Modern Inorganic Chemistry. New York: Longmans, Green, and Company.
• Petrucci, Ralph H.; Harwood, William S.; Herring, F. Geoffrey (2002). General Chemistry (8th ed.). Prentice-Hall. pp. 557–558. ISBN 0-13-014329-4.

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