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Freezing point depression occurs when there is non-volatile solute dissolved in a solvent. The presence of a solute decreases the temperature at which the solution freezes. The freezing point depression can be expressed in the terms of this equation:

ΔTf = Kf x mB

ΔTf is the change in freezing point temperature, Kf is the cryoscopic (freezing point depression) constant, and mB is the molality of solute (moles of solute per kilograms of solvent.)

As the molecules try to pack together to form a solid, the solute particles act as obstacles that block the solution molecules from packing. This requires that solution molecules to have even less kinetic energy (molecules becoming slower).

Freezing point depression is another example of a colligative property, meaning that it is affected only by the amount and not the identity of the solute. This makes sense because there is no sophisticated interaction between the solute particles and solvent particles; solute particles simply collides with solvent particles to prevent them from sticking to each other.

Freezing point depression is very useful for determining the molar mass of a compound and other things as well such as creating anti-freeze, melting snow on roads, and making ice cream. Sample Calculation:

Have you ever put salt on your skin and then put ice over it? I know, it hurts. The reason why it hurts your skin is because the salt makes the ice become colder than 0 ºC. We are going to measure how much colder the ice gets if you put a certain amount of salt over it. So lets see what happens when you put .25 grams of salt on 2 grams of ice.

ΔT = Kf x molality x i

What’s the i stand for? The vant’t Hoff factor, because salt is an ionic compound that dissociates in water. Since NaCl exists as Na+ and Cl ions in water, we have to consider them as two particles and i = 2.

And the Kf for water is 1.853 ºC ·kg/mol.

Let’s calculate the molality of the solution first.

We need to figure out the mols of solute to figure out the molality.

.25 g NaCl x 1mol / 58.4 g/mol = 0.00428 mols NaCl

We also need to figure out the kg of solvent.

2 g ice x 1kg / 1000g = 0.002 kg ice

0.00428 mols NaCl / 0.002 kg = 2.14 m

We plug all this in and we get,

ΔT = 1.853 ºC · kg/mol x 2 x .482 mol· kg

âˆ†T = 17.9 ºC