Animated Solution for Chemistry - Solutions: $K_f$ for water is $1.86{\text{ K kg mol}}^{-1}$. If your automobile radiator holds $1.0\text{ kg}$ of water, then how many grams of ethylene glycol (${\text{C}}_2{\text{H}}_6{\text{O}}_2$) must you add to get the freezing point of the solution lowered to $-2.8^{\circ }\text{C}$?

The problem of keeping a car engine from freezing in the dead of winter is a classic application of colligative properties. When water freezes, it expands, which can easily crack a solid metal engine block. To prevent this, we use a coolant—typically ethylene glycol—to lower the freezing point of the water inside the radiator. Let's dive into the beautiful chemistry that saves engines worldwide!

Imagine you are preparing your car for a harsh winter. Your radiator holds exactly $1.0\text{ kg}$ of water. Left alone, this pure water would freeze solid at $0^{\circ }\text{C}$.

Our goal is to prevent this disaster by adding ethylene glycol (${\text{C}}_2{\text{H}}_6{\text{O}}_2$) to lower the freezing point to a safe $-2.8^{\circ }\text{C}$. The question asks for the exact mass of ethylene glycol required to achieve this specific temperature drop.

How does adding a chemical prevent freezing? This phenomenon is governed by a colligative property known as the depression in freezing point. The fundamental equation is:

Here, $\Delta T_f$ is the change in the freezing point, $i$ is the van't Hoff factor, $K_f$ is the molal depression constant of the solvent, and $m$ is the molality of the solution.

Since ethylene glycol is a covalent organic molecule, it does not dissociate into ions when dissolved in water. Therefore, its van't Hoff factor is exactly $i=1$.

Let's gather our known values before we compute. The required drop in temperature, $\Delta T_f$, is the difference between the pure solvent's freezing point and the solution's freezing point:

Next, we calculate the molar mass of our solute, ethylene glycol (${\text{C}}_2{\text{H}}_6{\text{O}}_2$):

Our solvent, water, is already given at a convenient mass of $W_A=1.0\text{ kg}$. The molal depression constant for water is provided as $K_f=1.86{\text{ K kg mol}}^{-1}$.

Now, we bring it all together by substituting our values into the expanded molality equation. Remember, molality is the moles of solute divided by the mass of the solvent in kilograms:

Substituting the knowns:

To isolate the unknown mass $w_B$, we rearrange the equation:

The math simplifies beautifully:

So, to keep your engine safe, you need to pour in approximately $93\text{ g}$ of ethylene glycol.

Here is a fascinating thought experiment. What if we used table salt ($\text{NaCl}$) instead? It splits into two ions (${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$), giving an $i$ value of $2$. This means we would need half the mass to get the exact same freezing point drop!

But there is a catch. Saltwater is highly corrosive and would rapidly destroy the metal radiator. Ethylene glycol is non-corrosive and has the added benefit of raising the boiling point for summer driving. Chemistry isn't just about numbers; it's about choosing the right tool for the real world!