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

Freezing point depression occurs when a solute is added to a solvent, resulting in a lower freezing point for the solution compared to the pure solvent. The formula for calculating the change in freezing point is:

ΔT_f=i×K_f×molality where 'i' is the van't Hoff factor, 'K_f' is the freezing point depression constant specific to the solvent, and molality measures the moles of solute per kilogram of solvent. To find the freezing point of the solution, subtract ΔTf from the normal freezing point of the pure solvent. This concept is crucial in understanding how solutions behave differently from their pure components and has practical applications, such as in the formulation of antifreeze solutions.

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. Now, 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 ΔT_f=i×K_f×m which stands for Change in Freezing point equals i, which is your Van Hoff factor times K_f, 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 ΔT_f. If 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 the 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 110.7g of glucose, which is C₆H₁₂O₆ dissolved in 302.6g of water. All right, so we're going to say here that our freezing point of our solution equals 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°C. So what we need to do here is figure out what ΔT_F is.

We're going to say ΔT_F = i × K_F × m. Glucose is our solute. It is not ionic, it's covalent, so it doesn't break up into ions. So i is 1. K_F, we're dealing with water as a solvent. Its freezing point constant is 1.86°C/m. Then all we have to do now is find the moles of glucose and convert grams of water into kilograms of water. Kilograms just move the decimal over 3.

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 sit here for everyone. Mole of glucose. Its mass is 180.156g, and that's the weight of the six carbons, 12 hydrogens and six oxygens. Together 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. So here these cancel out. With this molality here, what I get at the end is 3.78°C, which is what I plug over here. So the new freezing point after the additional glucose is -3.78 degrees Celsius. So that would 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 2.89°C. 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?

Freezing point depression is a colligative property of solutions, which means that it depends on the ratio of solute particles to solvent molecules in a solution, not on the identity of the solute. Specifically, it refers to the phenomenon where the freezing point of a liquid (a solvent) is lowered by adding a solute to it. This happens because the solute particles disrupt the formation of the crystalline structure of the solvent needed for it to freeze, thereby requiring a lower temperature to achieve the same structure.

For example, when salt is added to ice, the freezing point of the water is lowered, which is why salt is often used to melt ice on roads in winter. The extent of freezing point depression can be calculated using the formula:

ΔT f = i ⋅ K f ⋅ m where ΔTf is the freezing point depression, i is the van 't Hoff factor (which represents the number of particles the solute breaks into in solution), Kf is the freezing point depression constant of the solvent, and m is the molality of the solution.

How do you calculate freezing point depression?

Freezing point depression is calculated using the formula:

ΔT f = i ⋅ K f ⋅ m where ΔT_f is the freezing point depression, i is the van't Hoff factor (which represents the number of particles the solute splits into or forms in solution), K_f is the cryoscopic constant (freezing point depression constant) of the solvent, and m is the molality of the solution.

Here's how you use the formula:

1. Determine the cryoscopic constant (K_f) for the solvent, which is a known value that you can usually find in a reference book or online.
2. Calculate the molality (m) of the solution, which is the number of moles of solute per kilogram of solvent.
3. Identify the van't Hoff factor (i) for the solute, which is typically 1 for non-electrolytes (substances that don't dissociate in solution) and more than 1 for electrolytes (substances that dissociate into ions).
4. Multiply these values together to find the freezing point depression (ΔT_f).

The freezing point of the solution is then found by subtracting ΔT_f from the pure solvent's normal freezing point.

Why does freezing point depression occur?

Freezing point depression occurs due to the presence of solute particles in a solvent. When a solute, such as salt, is dissolved in a solvent, like water, it disrupts the formation of the solvent's solid phase (ice in the case of water). This disruption happens because the solute particles interfere with the ability of the solvent molecules to come together and form a regular crystal lattice, which is necessary for freezing.

The presence of solute particles means that the solvent molecules must lose more kinetic energy than they would have to if the solute were not present, in order to form the orderly crystal structure of ice. This means that the temperature must be lowered further than the pure solvent's freezing point to achieve freezing. The more solute particles that are present, the greater the disruption to the crystal formation, and thus, the lower the temperature needs to be to freeze the solvent. This phenomenon is a colligative property, meaning it depends on the number of solute particles in the solution, not their identity.