In general, solutions have a lower freezing point than does the pure solvent. This freezing point depression is an example of a colligative property, one which depends only on the number of moles of the solute. not the identity of that solute. This is a useful property in many ways: for example, car antifreeze is a mixture of water and ethylene glycol which has a freezing point much lower than that of pure water.

The amount of the depression is given by

DT_f = k_fm

When working with ionic solutes in water and other polar solvents, one must be careful to take into consideration the fact that the ion concentration is higher than the concentration of the solute. For example, table salt, NaCl dissolves in water to form ions

NaCl(s) -> Na₊(aq) + Cl^-(aq)

Example 1: The normal freezing point of benzene is 5.5 ^oC. It has a freezing point depression constant of -4.90 ^oC/m. If we make up a 0.500 molal solution of Br₂ in benzene, what is the freezing point of the mixture?

Solution: Simply use the above equation and substitute in the proper values

DT_f = k_fm

DT_f = -4.90 ^oC/m * 0.500 molal

DT_f = -2.45 ^oC

Example 2: The normal freezing point of water is 0.00 ^oC and it's freezing point depression constant is -1.86 ^oC/m. If we make up a 1.00 molal solution of Na₂SO₄, what is the freezing point of the mixture.

Solution 2: Again, use the above equation but we need to be careful to account for the ions. Na₂SO₄ fragments into three pieces when it dissolves,

Na₂SO₄(s) -> 2Na⁺(aq) + SO₄^-2(aq)

DT_f = k_fm

DT_f = -1.86 ^oC/m * 3.00 molal

DT_f = -5.58 ^oC

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