Freezing Point Depression - Online Tutor, Practice Problems & Exam Prep

Freezing point depression occurs when a solute is added to a pure solvent, lowering its freezing point. The change in freezing point (ΔT_f) can be calculated using the formula ΔT_f = I × K_f × m, where I is the van’t Hoff factor, K_f is the freezing point constant, and m is the molality (moles of solute per kilogram of solvent). The freezing point of the solution is the normal freezing point of the solvent minus ΔT_f. More solute results in a greater decrease in freezing point.

Freezing Point Depression Concept 1

The phenomenon when adding a solute to a pure solvent results in the decreasing of the freezing point of the solvent. Now with this whole idea of freezing point depression, we have what are called the normal freezing point and the freezing point of the solution. The normal freezing point, which I'm going to abbreviate as fp, has to do with our pure solvent. We're going to say this is the freezing point of the solvent before the addition of the solute. And the freezing point of the solution, which we're going to say is fp solution, this is the freezing point of the solvent after the addition of solute. Remember, once you add solute to solvent, it becomes a solution. Now if we take a look at freezing point depression, let's look at some important areas.

First, we have the freezing point depression formula, which is Δtf=I⋅kf⋅molality, which stands for our change in freezing point, equals I, which is your van't Hoff factor, times kf, which is your freezing point constant of the solvent in degrees Celsius over molality, times molality itself. Remember, molality is just moles of solute over kilograms of solvent. Now that we know the components of the freezing point depression formula, realize how does it relate to the freezing point of a solution? Well, we're going to say freezing point of a solution equals freezing point of the pure solvent minus Δtf. And we take a look here, we can see that we have some common types of solvents. We have water, benzene, chloroform, and ethanol, each with their own normal freezing point before the addition of any solute. Each of them also has their own unique freezing point constant value. No, you don't need to memorize these. This is just for reference of some of the most common types of solvents you may see. So just remember, when it comes to freezing point depression, our freezing point decreases the more solute we add to our solvent.

Freezing Point Depression Example 1

Calculate the freezing point of a solution containing 110.7 grams of glucose, which is C₆H₁₂O₆, dissolved in 302.6 grams of water. Alright. So we're going to say here that the freezing point of our solution equals the freezing point of our solvent minus ΔT_f. Our solvent here is water; pure water freezes at 0 degrees Celsius. So what we need to do here is figure out what ΔT_f is. We're going to say ΔT_f equals I × K_f × m.

Glucose here is our solute. It is not ionic, it's covalent, so it doesn't break up into ions. Thus, I is 1. K_f, as we're dealing with water as a solvent, its freezing point constant is 1.86 degrees Celsius per molality. Then all we have to do now is find the moles of glucose and convert grams of water into kilograms of water. For kilograms, we just move the decimal over 3 places, so there goes our kilograms of our solvent water. And here we're going to find out the moles of our glucose. So we have this many grams of glucose. We're going to say here that for every 1 mole of glucose, its mass is 180.156 grams. This is the weight of the 6 carbons, 12 hydrogens, and 6 oxygens combined. Grams cancel out, and now I'm going to have moles which comes up to 0.614467 moles of glucose. So we plug that here. These cancel out with this molality here. What I get at the end is 3.78 degrees Celsius, which is what I plug over here. So the new freezing point after the addition of glucose is negative 3.78 degrees Celsius. So that will be our final answer.

How many moles of ethylene glycol, C₂H₆O₂, must be added to 1,000 g of water to form a solution that has a freezing point of – 10ºC?

• A. 334 moles
• B. 5.4 moles
• C. 3.2 moles
• D. 200 moles

An ethylene glycol solution contains 28.3 g of ethylene glycol, C₂H₆O₂ in 97.2 mL of water. Calculate the freezing point of the solution. The density of water 1.00 g/mL.

• A. –8.72°C
• B. –0.848°C
• C. –0.541°C
• D. –17.4°C

When 825 g of an unknown is dissolved in 3.45 L of water, the freezing point of the solution is decreased by 2.89ºC. Assuming that the unknown compound is a non-electrolyte, calculate its molar mass.

• A. 154 g/mol
• B. 42.4 g/mol
• C. 44.5 g/mol
• D. 159 g/mol

Here’s what students ask on this topic:

What is freezing point depression and how is it calculated?

Freezing point depression occurs when a solute is added to a pure solvent, causing the freezing point of the solvent to decrease. The change in freezing point (ΔT_f) can be calculated using the formula:

$\mathrm{\Delta T_f}=I\times \mathrm{K_f}\times m$

where I is the van’t Hoff factor (number of particles the solute dissociates into), K_f is the freezing point depression constant of the solvent, and m is the molality (moles of solute per kilogram of solvent). The freezing point of the solution is then the normal freezing point of the pure solvent minus ΔT_f.

How does the van’t Hoff factor affect freezing point depression?

The van’t Hoff factor (I) represents the number of particles into which a solute dissociates in solution. It directly affects the magnitude of freezing point depression. For example, if a solute dissociates into more particles, the van’t Hoff factor increases, leading to a greater decrease in the freezing point. The formula for freezing point depression is:

$\mathrm{\Delta T_f}=I\times \mathrm{K_f}\times m$

Thus, a higher I value results in a larger ΔT_f, meaning the solution's freezing point is lower.

What is the difference between the normal freezing point and the freezing point of a solution?

The normal freezing point is the temperature at which a pure solvent freezes. It is denoted as f.p. The freezing point of a solution, on the other hand, is the temperature at which a solvent containing a dissolved solute freezes. This temperature is lower than the normal freezing point due to freezing point depression. The relationship can be expressed as:

$\mathrm{f.p.\; solution}=\mathrm{f.p.\; pure\; solvent}-\mathrm{\Delta T_f}$

where ΔT_f is the change in freezing point caused by the addition of the solute.

Why does adding a solute to a solvent lower its freezing point?

Adding a solute to a solvent lowers its freezing point because the solute particles disrupt the formation of the solid structure of the solvent. This disruption requires a lower temperature to achieve the same solid structure, thus lowering the freezing point. The phenomenon is quantified by the freezing point depression formula:

$\mathrm{\Delta T_f}=I\times \mathrm{K_f}\times m$

where I is the van’t Hoff factor, K_f is the freezing point depression constant, and m is the molality of the solution.

What are some common solvents and their freezing point constants?

Some common solvents and their freezing point constants (K_f) are:

• Water: K_f = 1.86 °C·kg/mol
• Benzene: K_f = 5.12 °C·kg/mol
• Chloroform: K_f = 4.68 °C·kg/mol
• Ethanol: K_f = 1.99 °C·kg/mol

These constants are used in the freezing point depression formula to calculate the change in freezing point when a solute is added to these solvents.