<meta http-equiv="refresh" content="1; url=/nojavascript/"> Freezing Point Depression ( Read ) | Chemistry | CK-12 Foundation
Skip Navigation
You are viewing an older version of this Concept. Go to the latest version.

Freezing Point Depression

Best Score
Practice Freezing Point Depression
Best Score
Practice Now
Freezing Point Depression
 0  0  0

Why salt icy roads?

Colligative properties have practical applications, such as the salting of roads in cold-weather climates. By applying salt to an icy road, the melting point of the ice is decreased, and the ice will melt more quickly, making driving safer. Sodium chloride (NaCl) and either calcium chloride (CaCl 2 ) or magnesium chloride (MgCl 2 ) are used most frequently, either alone or in a mixture. Sodium chloride is the least expensive option, but is less effective because it only dissociates into two ions instead of three.

Freezing Point Depression

The Figure below shows the phase diagram for a pure solvent and how it changes when a solute is added to it. The solute lowers the vapor pressure of the solvent resulting in a lowering of the freezing point of the solution compared to the solvent. The freezing point depression is the difference in temperature between the freezing point of the pure solvent and that of the solution. On the graph, the freezing point depression is represented by \Delta T_f .

The vapor pressure of a solution (blue) is lower than the vapor pressure of a pure solvent (pink). As a result, the freezing point of a solvent decreases when any solute is dissolved into it.

When a pure solvent freezes, its particles become more ordered as the intermolecular forces that operate between the molecules become permanent. In the case of water, the hydrogen bonds make the hexagonally-shaped network of molecules that characterizes the structure of ice. By dissolving a solute into the liquid solvent, this ordering process is disrupted. As a result, more energy must be removed from the solution in order to freeze it, and the freezing point of the solution is lower than that of the pure solvent. 

The magnitude of the freezing point depression is directly proportional to the molality of the solution. The equation is:

\Delta T_f=K_f \times m

The proportionality constant, K_f , is called the molal freezing-point depression constant . It is a constant that is equal to the change in the freezing point for a 1-molal solution of a nonvolatile molecular solute. For water, the value of  K_f is -1.86°C/ m . So the freezing temperature of a 1-molal aqueous solution of any nonvolatile molecular solute is -1.86°C. Every solvent has a unique molal freezing-point depression constant. These are shown in Table below , along with a related value for the boiling point called K_b .

Molal Freezing-Point and Boiling-Point Constants
Solvent Normal freezing point (°C) Molal freezing-point depression constant, K f (°C/ m ) Normal boiling point (°C) Molal boiling-point elevation constant, Kb (°C/ m )
Acetic acid 16.6 -3.90 117.9 3.07
Camphor 178.8 -39.7 207.4 5.61
Naphthalene 80.2 -6.94 217.7 5.80
Phenol 40.9 -7.40 181.8 3.60
Water 0.00 -1.86 100.00 0.512

Sample Problem: Freezing Point of a Nonelectrolyte

Ethylene glycol (C 2 H 6 O 2 ) is a molecular compound that is used in many commercial anti-freezes. A water solution of ethylene glycol is used in vehicle radiators to lower its freezing point and thus prevent the water in the radiator from freezing. Calculate the freezing point of a solution of 400. g of ethylene glycol in 500. g of water.

Step 1: List the known quantities and plan the problem.


  • mass C 2 H 6 O 2 = 400. g
  • molar mass C 2 H 6 O 2 = 62.08 g/mol
  • mass H 2 O = 500.0 g = 0.500 kg
  • \text{mass H}_2\text{O}=500.0 \ \text{g}=0.500 \ \text{kg}
  • K_f(\text{H}_2\text{O})=-1.86^\circ \text{C}/m


  • T_f \ \text{of solution}=? \ ^\circ \text{C}

This is a three-step problem. First, calculate the moles of ethylene glycol. Then, calculate the molality of the solution. Finally, calculate the freezing point depression.

Step 2: Solve.

400. \text{ g C}_2\text{H}_6\text{O}_2 \times \frac{1 \text{ mol C}_2\text{H}_6\text{O}_2}{62.08 \text{ g C}_2\text{H}_6\text{O}_2} &= 6.44 \text{ mol C}_2\text{H}_6\text{O}_2\\\frac{6.44 \text{ mol C}_2\text{H}_6\text{O}_2}{0.500 \text{ kg H}_2\text{O}} &= 12.9 \ m \ \text{C}_2\text{H}_6\text{O}_2\\\Delta T_f=K_f \times m=-1.86^\circ \text{C/}m \times 12.9 \ m &= -24.0^\circ \text{C} \\T_f &=-24.0^\circ \text{C}

The normal freezing point of water is 0.0°C. Therefore, since the freezing point decreases by 24.0°C, the freezing point of the solution is -24.0°C.

Step 3: Think about your result.

The freezing point of the water decreases by a large amount, protecting the radiator from damage due to the expansion of water when it freezes. There are three significant figures in the result.


  • Freezing point depression is defined.
  • Calculations involving freezing point depression are described.


Solve the problems at the site below:



  1. How does a solute affect the freezing of water?
  2. How many moles of glucose would be needed to lower the freezing point of one kg of water 3.72°C?
  3. How many moles of NaCl would be needed to produce the same amount of lowering of temperature?

Image Attributions


Email Verified
Well done! You've successfully verified the email address .
Please wait...
Please wait...
ShareThis Copy and Paste

Original text