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(Video) Ice Cream and Freezing Point Depression

by HarvardX

Have you ever eaten ice cream that had huge ice crystals in it? Why does some ice cream freeze like that while other kinds are velvety and smooth? The answer lies in freezing point depression.

When you add solvents like sugar or starches to a milk base, it lowers the freezing point of that solution. In other words, you can't freeze ice cream at the same temperature as you can pure water. The video above includes a great diagram to show why this is the case.

This is also the reason why some ice cream has smaller ice crystals. It's because the water molecules in the ice cream are more separated and can't form large chunks when they freeze.

Do you live in a snowy place? I do. In Colorado, we get tons of snowstorms. Each storm, the mountains can get feet of snow! The worst part is our major interstate goes right through the mountains. Driving becomes unbelievably dangerous! But, it's great for skiing and snowboarding.

Did you know, before snowstorms, they put magnesium chloride ($MgCl_{2}$) solution on the roads? But, why? The $MgCl_{2}$ lowers the water's freezing temperature! So, ice doesn't form on roads. We call this freezing point depression.

Colligative Properties

Freezing point depression is an example of a colligative property. A solution's colligative properties are physical properties that depend on the solute's amount. It does not matter what solute we use!

For colligative properties, the solvent is always our reference point. Let's consider water, the universal solvent. When we add $MgCl_{2}$ to water, we lower water's freezing point ($0^{o}C$). How much do we depress the freezing point? It depends on the solution's molality. What happens if we use sodium chloride ($NaCl$) instead? Equal molalities mean the freezing point depressions are the same.

Molality?!!! Why?

Are you wondering why we use molality? Remember, molality is moles of solute per kilogram of solvent ($\frac{mol_{solute}}{kg_{solvent}}$). Why don't we use molar concentrations? Molarity is moles of solute per liter of solvent ($\frac{mols}{L}$). The bottom term is volume. But, volumes are temperature-dependent. We know this is true for gases. Just look at Charles' law or the ideal gas law. It's also true for liquids and solids. But, those equations are complicated. Molality depends on mass. Mass is never temperature-dependent. So, we can use it despite the temperature change.

Freezing Point Depression Formula

To help you understand, let's look at the freezing point depression formula. The equation is $\Delta T_{f}=K_{f}\times m$.

• $\Delta T_{f}$ is the change in the freezing point.

• $K_{f}$ is the freezing point depression constant.

• Its units are $\frac{^{o}C}{m}$.

• The constant is unique for each solvent. The most common solvent is water. Its $K_{f}=1.86\frac{^{o}C}{m}$.

• $m$ is the solution's molality.

• Remember molality's units are $\frac{mol_{solute}}{kg_{solvent}}$.

Why does Freezing Point Depression Happen?

Do you remember our phase change diagrams? A phase change occurs at a phase boundary. The phase boundaries are the temperatures and pressures that create an equilibrium. So, we're shifting the phase equilibrium temperatures.

We know that equilibrium constants are temperature-dependent. They also relate to the Gibbs free energy. With a pure solvent, we only have interactions between solvent molecules. When we add a solute, we add solute-solvent interactions. So, we increased the entropy. Our energy is more spread out. Equilibrium is when the Gibbs free energy equals zero. The solute addition pushed Gibbs free energy further from zero. So, we have to lower the temperature to reach the phase change equilibrium.

Freezing Point Depression Example Problem

Our cars use freezing point depression every day. We mix water with ethylene glycol to make antifreeze. The antifreeze circulates through the coolant system. The engine transfers heat to the antifreeze. So, your car's engine stays at safe operating temperatures. Let's look at an example problem based on antifreeze.

Let's say we mixed one gallon of antifreeze solution (about 3.8 liters). We mix ethylene glycol with water in a 1:1 ratio. The density of ethylene glycol is $0.7857\frac{g}{ml}$. Its molar mass is $62.1\frac{g}{mol}$. Calculate the freezing point depression of water.

Step 1: Find the kilograms of water and moles of ethylene glycol. Use dimensional analysis.

• $3.8 L_{solution} \times \frac{1 L_{ethylene\,glycol}}{2L_{solution}} \times \frac{1000.0 ml}{L} \times \frac{0.7857g}{ml} \times \frac{1mol}{62.1g}=24.0 mols_{ethlyene\,glycol}$

• $3.8 L_{solution}\times \frac{1.0 L_{water}}{2.0 L_{solution}} \times\frac{1.0kg}{L}=1.9\,kg_{water}$

Step 2: Calculate the molality.

• $m=\frac{mol_{solute}}{kg_{solvent}}=\frac{24.0mols_{ethylene\,glycol}}{1.9kg_{water}}= 12.6 molal$

Step 3: Calculate the freezing point depression.

• $\Delta T_{f}=K_{f}\times m \rightarrow 1.86\frac{^{o}C}{m}\times 12.6m=23.5^{o}C$

• We lowered (depressed) the freezing point of water by $23.5^{o}C$. So the new freezing point would be $-23.5^{o}C$ or $-10.0^{o}F$.

• What if you lived somewhere that consistently got below these temperatures? You would need more ethylene glycol in your antifreeze.