The freezing point of a solution is lower than that of the solvent from which it is formed.

Freezing point depression is a colligative property of solutions. Solutions freezing points are lower than that of the pure solvent or solute because freezing, or becoming solid, creates order and decreases entropy. Solutions have high entropy because of the mix of solvent and solute, so it takes more energy to decrease their entropy to the same point.

freezing point depression elevation pressure vapor

Alright when dealing with solutions, you're going to be coming across a colligative properties and one of the colligative properties that you're going to see is freezing point depression and that says in a solution solute particles interfere with attractive forces among the solvent particles. And this prevents solution into entering a solid state. So essentially what they're saying is, because a liquid has like all these extra particles in it to make a solution, and it's not a pure solvent those get in the way of the intermolecular forces that make it a solid, a solid. Like the hydrogen bonding, dipole-dipole interaction and the London dispersion forces, those extra particles that are there kind of get in the way and they actually help to like lower the freezing point to get, to push those particles out so it'll be a pure solvent when it's actually frozen. Freezing point states that the particles are no longer have sufficient kinetic energy to overcome intermolecular attractive forces so when those particles are there those attractive forces are not necessary. So like they're going to have to just push those particles out so they can actually have the inner particle attractive forces present. So let's actually talk about different substances and their freezing points. So we're talking about water which is a universal solvent and we know that water freezes typically at 0 degrees Celsius at normal freezing point. Now for every molal of substance I'm going to put within that water in a pure substance is actually going to drop the temperature the freezing point temperature by 1.86 degrees Celsius.Benzene freezes at 5.5 degrees Celsius well higher than water and for every mol of substance that you have for a kilogram of solution the boiling point is going to drop even more 5.12 degrees Celsius and so on and so forth. So if you were to look at this, this is very similar to boiling point elevation but this set formula is exactly the same but there's some slight differences. So the change in temperature of the freezing point is equal to the constant that we had discussed, time similarity of this solution that we're dealing with, time is a Ben Hoff Factor and a Ben Hoff Factor is how much the particle actually separates in solution. So we're talking about ionic compounds, they separate into solution depending on how many particles they have or how many ions they have in that but molecular compounds don't at all. So let's actually put that in action.Alright so what a freezing point of a 0.029 molal of NaCl aqueous solution so we know it's aqueous and the aqueous tells us that our solvent is water. So we're going to say our delta T are changed in temperature to freezing point is going to equal to the constant of water which is 1.86 degrees Celsius for every molar. And the molar solution is 0.029 and because it's NaCl I know it's ionic for every one molal it's going to actually separate into 2 substances Na plus and Cl minus. So we're actually multiplying this by 2, we have substances when it's in solution. So when you multiply all of these together you get 0.11 degrees Celsius and we're going to say alright our original freezing point is 0 it's going to lower by 0.11 and so our new freezing point is 0.11 degrees Celsius negative because it dropped that much. So we can actually like talk about, when you think about when it snows outside and the reason that you put salt on the roads there isn't even salt on the roads that's actually going to lower the freezing point so there's not going to be sheets of ice on your drive way or on the roads or on the side walks. So that's why they use salt and they actually use calcium chloride typically which is actually better than sodium chloride because this actually breaks up into 3 particles so it'll drop the freezing point 3 times much as another solute would. So this is an example of frizzing point depression.

Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent. The freezing points of solutions are all lower than that of the pure solvent and is directly proportional to the molality of the solute.

\[\Delta{T_f} = T_f(solvent) - T_f (solution) = K_f \times m\]

where \(\Delta{T_f}\) is the freezing point depression, \(T_f\) (solution) is the freezing point of the solution, \(T_f\) (solvent) is the freezing point of the solvent, \(K_f\) is the freezing point depression constant, and m is the molality.

Nonelectrolytes are substances with no ions, only molecules. Strong electrolytes, on the other hand, are composed mostly of ionic compounds, and essentially all soluble ionic compounds form electrolytes. Therefore, if we can establish that the substance that we are working with is uniform and is not ionic, it is safe to assume that we are working with a nonelectrolyte, and we may attempt to solve this problem using our formulas. This will most likely be the case for all problems you encounter related to freezing point depression and boiling point elevation in this course, but it is a good idea to keep an eye out for ions. It is worth mentioning that these equations work for both volatile and nonvolatile solutions. This means that for the sake of determining freezing point depression or boiling point elevation, the vapor pressure does not effect the change in temperature. Also, remember that a pure solvent is a solution that has had nothing extra added to it or dissolved in it. We will be comparing the properties of that pure solvent with its new properties when added to a solution.

Adding solutes to an ideal solution results in a positive ΔS, an increase in entropy. Because of this, the newly altered solution's chemical and physical properties will also change. The properties that undergo changes due to the addition of solutes to a solvent are known as colligative properties. These properties are dependent on the number of solutes added, not on their identity. Two examples of colligative properties are boiling point and freezing point: due to the addition of solutes, the boiling point tends to increase, and freezing point tends to decrease.

The freezing point and boiling point of a pure solvent can be changed when added to a solution. When this occurs, the freezing point of the pure solvent may become lower, and the boiling point may become higher. The extent to which these changes occur can be found using the formulas:

\[\Delta{T}_f = -K_f \times m\]

\[\Delta{T}_f = K_b \times m\]

where \(m\) is the solute molality and \(K\) values are proportionality constants; (\(K_f\) and \(K_b\) for freezing and boiling, respectively.

If solving for the proportionality constant is not the ultimate goal of the problem, these values will most likely be given. Some common values for \(K_f\) and \(K_b\) respectively, are:

Molality is defined as the number of moles of solute per kilogram solvent. Be careful not to use the mass of the entire solution. Often, the problem will give you the change in temperature and the proportionality constant, and you must find the molality first in order to get your final answer.

The solute, in order for it to exert any change on colligative properties, must fulfill two conditions. First, it must not contribute to the vapor pressure of the solution, and second, it must remain suspended in the solution even during phase changes. Because the solvent is no longer pure with the addition of solutes, we can say that the chemical potential of the solvent is lower. Chemical potential is the molar Gibb's energy that one mole of solvent is able to contribute to a mixture. The higher the chemical potential of a solvent is, the more it is able to drive the reaction forward. Consequently, solvents with higher chemical potentials will also have higher vapor pressures.

The boiling point is reached when the chemical potential of the pure solvent, a liquid, reaches that of the chemical potential of pure vapor. Because of the decrease in the chemical potential of mixed solvents and solutes, we observe this intersection at higher temperatures. In other words, the boiling point of the impure solvent will be at a higher temperature than that of the pure liquid solvent. Thus, boiling point elevation occurs with a temperature increase that is quantified using

\[\Delta{T_b} = K_b b_B\]

where

• \(K_b\) is known as the ebullioscopic constant and
• \(m\) is the molality of the solute.

Freezing point is reached when the chemical potential of the pure liquid solvent reaches that of the pure solid solvent. Again, since we are dealing with mixtures with decreased chemical potential, we expect the freezing point to change. Unlike the boiling point, the chemical potential of the impure solvent requires a colder temperature for it to reach the chemical potential of the pure solid solvent. Therefore, a freezing point depression is observed.

Example \(\PageIndex{1}\)

2.00 g of some unknown compound reduces the freezing point of 75.00 g of benzene from 5.53 to 4.90 \(^{\circ}C\). What is the molar mass of the compound?

Solution

First we must compute the molality of the benzene solution, which will allow us to find the number of moles of solute dissolved.

\[ \begin{align*} m &= \dfrac{\Delta{T}_f}{-K_f} \\[4pt] &= \dfrac{(4.90 - 5.53)^{\circ}C}{-5.12^{\circ}C / m} \\[4pt] &= 0.123 m \end{align*}\]

\[ \begin{align*} \text{Amount Solute} &= 0.07500 \; kg \; benzene \times \dfrac{0.123 \; m}{1 \; kg \; benzene} \\[4pt] &= 0.00923 \; m \; solute \end{align*}\]

We can now find the molecular weight of the unknown compound:

\[ \begin{align*} \text{Molecular Weight} =& \dfrac{2.00 \; g \; unknown}{0.00923 \; mol} \\[4pt] &= 216.80 \; g/mol \end{align*}\]

The freezing point depression is especially vital to aquatic life. Since saltwater will freeze at colder temperatures, organisms can survive in these bodies of water.

Road salting takes advantage of this effect to lower the freezing point of the ice it is placed on. Lowering the freezing point allows the street ice to melt at lower temperatures. The maximum depression of the freezing point is about −18 °C (0 °F), so if the ambient temperature is lower, \(\ce{NaCl}\) will be ineffective. Under these conditions, \(\ce{CaCl_2}\) can be used since it dissolves to make three ions instead of two for \(\ce{NaCl}\).

Benzophenone has a freezing point of 49.00oC. A 0.450 molal solution of urea in this solvent has a freezing point of 44.59oC. Find the freezing point depression constant for the solvent.(answ.: 9.80oC/m)