Freezing Point Depression Formula and Definition

Freezing Point Depression
Freezing point depression is the lowering of freezing point caused by dissolving another substance in a liquid.

Freezing point depression is lowering the temperature of the freezing point of a liquid by dissolving another substance in it. Like boiling point elevation and osmotic pressure, it is a colligative property of matter.

How Freezing Point Depression Works

What this means is that the amount of freezing point depression depends on how many particles dissolve in the liquid, not on their chemical identity. So, the freezing point depression from dissolving salt (NaCl) in water is greater than the effect of dissolving sugar in water (C12H22O11) because each salt molecule dissociates into two particles (Na+ and Cl ions), while sugar dissolves but does not dissociate. Calcium chloride (CaCl2) depresses freezing point more than table salt because it dissociates into three particles in water (one Ca+ and two Cl ions).

In general electrolytes cause a greater freezing point depression than nonelectrolytes. But, solubility in a solvent also matters. So, salt (NaCl) produces a larger freezing point depression in water than magnesium fluoride (MgF2). Even though magnesium fluoride dissociates into three particles and salt dissociates into three particles, magnesium fluoride is insoluble in water.

The reason the number of particles makes a difference is because these particles get between solvent molecules and disrupt the organization and bond formation that causes liquids to freeze or solidify.

Freezing Point Depression Examples

Freezing point depression occurs in everyday life. Here are some examples.

  • The freezing point of sea water is lower than that of pure water. Sea water contains numerous dissolved salts. One consequence of this is that rivers and lakes often freeze in the winter when temperatures drop below 0 °C. It takes much colder temperatures to freeze the ocean.
  • When you put salt on an icy walk, freezing point depression prevents melting ice from re-freezing.
  • Adding salt to ice water lowers its temperature enough that you can make ice cream without a freezer. All you do is place a sealed bag of ice cream mixture into a bowl of salted ice.
  • Antifreeze lowers the freezing point of water, keeping it from freezing in vehicles in the winter.
  • Vodka and other high-proof alcoholic beverages do not freeze in a home freezer. Alcohol causes significant freezing point depression of water. However, the freezing point of vodka is higher than that of pure alcohol. So, take care to look at the freezing point of the solvent (water) and not the solute (ethanol) in freezing point depression calculations!

Freezing Point Depression Formula

The freezing point depression formula uses the Clausius-Clapeyron equation and Raoult’s law. For a dilute ideal solution, the formula for freezing point depression is called Blagden’s law:

ΔTf = iKfm

  • ΔTf is the temperature difference between normal freezing point and the new freezing point
  • i is the van’t Hoff factor, which is the number of particles the solute breaks into
  • Kf is the molal freezing point depression constant or cryoscopic constant
  • m is the molality of the solution

The cryoscopic constant is a characteristic of the solvent, not the solute. This table lists Kf values for common solvents.

CompoundFreezing Point (°C)Kf in K·kg/mol
Acetic acid16.63.90
Carbon disulfide-1123.8
Carbon tetrachloride-2330
Ethyl ether-116.21.79
Cryoscopic constant or molal freezing point depression constant (Kf) for common solvents.

How to Calculate Freezing Point Depression – Example Problems

Note the freezing point depression formula only works in dilute solutions where the solute is present in much lower amounts than the solvent and when the solute is non-volatile.

Example #1

What is the freezing point of an aqueous solution of NaCl with a concentration of 0.25 m? The Kf of water is 1.86 °C/m.

In this case, i is 2 because salt dissociates into 2 ions in water.

ΔT = iKfm = (2)(1.86 °C/m)( 0.25 m) = 0.93 °C.

So, this means the freezing point of the solution is 0.93 degrees lower than the normal freezing point of water (0 °C). The new freezing point is 0 – 0.93 = -0.93 °C.

Example #2

What is the freezing point of water when 31.65 grams of sodium chloride (NaCl) dissolves in 220.0 mL of water at 35 °C. Assume the sodium chloride completely dissolves and that the density of water at 35 °C is 0.994 g/mL. The Kf for water is 1.86 °C · kg/mol.

First, find the molality (m) of the salt water. Molality is the number of moles of NaCl per kilogram of 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

Next, determine the van’t Hoff factor. For substances that don’t dissociate, like sugar, the van’t Hoff factor is 1. Salt dissociates into two ions: Na+ and Cl. So the van’t Hoff factor i is 2.

Now, we have all the information and can calculate ΔT.

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

Adding 31.65 g of NaCl to 220.0 mL of water will lower the freezing point by 9.21 °C. The normal freezing point of water is 0 °C, so the new freezing point is 0 – 9.21 or -9.21 °C.

Example #3

What is the freezing point depression when you dissolve 62.2 grams of toluene (C7H8) in 481 grams of naphthalene? The freezing point depression constant Kf for naphthalene is 7 °C · kg/mol.

First, calculate the molality of the solution. Toluene is an organic solute that does not dissociate into ions, so molality is the same as molarity.

m = 62.2 g / 92.1402 g/mol = 0.675058 m

Because toluene does not dissociate, its van’t Hoff factor is 1.

ΔT = iKfm = Kfm = (7.00 °C kg mol¯1) (0.675058 mol / 0.481 kg) = 9.82 °C

So, the freezing point depression is 9.82 degrees. Remember, this is the amount the freezing point lowers and not the new freezing point.


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