# What is the freezing point of an aqueous solution containing "40.0 g" of ethylene glycol in "60.0 g" of water? K_f = 1.86^@"C/m" and K_b = 0.512^@ "C/m" for water.

## A) ${20.1}^{\circ} \text{C}$ B) $- {20.1}^{\circ} \text{C}$ C) ${5.53}^{\circ} \text{C}$ D) $- {5.53}^{\circ} \text{C}$

Aug 8, 2016

I got $B$, $\approx - {20.1}^{\circ} \text{C}$.

As a note,

• $A$ is due to getting the wrong sign on $\Delta {T}_{f}$.
• $C$ is due to using ${K}_{b}$ instead of ${K}_{f}$ and getting the wrong sign.
• $D$ is due to using ${K}_{b}$ instead of ${K}_{f}$ and getting the right sign.

INITIAL PREDICTIONS

First off, let's predict the freezing point from thinking about this qualitatively. Pure ethylene glycol has a freezing point of $- {12.9}^{\circ} \text{C}$, and water's freezing point is ${0}^{\circ} \text{C}$.

So, the solution's freezing point should actually be below $\setminus m a t h b f \left({0}^{\circ} \text{C}\right)$ (what occurs is freezing point depression due to colligative properties of adding solutes into a solvent, so the freezing point should drop).

We can eliminate all but $\text{B}$, $\setminus m a t h b f \left(- {20.1}^{\circ} \text{C}\right)$. That should be our answer before doing any work at all.

FREEZING POINT DEPRESSION FORMULA

Now let's actually calculate it so we can prove it. The High School version of the formula for freezing point depression is:

$\setminus m a t h b f \left(\Delta {T}_{f} = {T}_{f} - {T}_{f}^{\text{*}} = {K}_{f} \cdot m \cdot i\right)$

where:

• ${T}_{f}$ is the freezing point of the solution.
• ${T}_{f}^{\text{*}}$ is the freezing point of the pure solvent.
• ${K}_{f}$ is the freezing point depression constant of the pure solvent.
• $m$ is the molal concentration of the solution, which is the $\text{mol}$s of solute per $\text{kg}$ of the pure solvent.
• $i$ is the van't Hoff factor, which for ideal solutions is equal to the number of ions that dissociate in solution per formula unit.

FREEZING POINT CALCULATION

Ethylene glycol is otherwise known as ethanediol. In this solution, we have:

"40.0" cancel("g") xx "1 mol"/("62.0" cancel("g")) = color(green)("0.6452 mols") ethanediol

$\text{60.0" cancel("g") xx "1 mol"/("18.015" cancel("g")) = "2.775 mols}$ water

Since we clearly have more water than ethanediol, it's safe to say that water is the solvent.

Therefore, we can use the $\textcolor{g r e e n}{{K}_{f}}$ of water, $\textcolor{g r e e n}{\text{1.86"^@"C/m}}$ (something you should be able to look up or have access to). Then, the molality $m$ of the solution is:

$\textcolor{g r e e n}{m} = \text{mols ethanediol"/"kg water}$

$= \text{0.6452 mols ethanediol"/"0.0600 kg water}$

$=$ $\textcolor{g r e e n}{\text{10.75 m}}$ (man, that is high!)

Also, ethanediol is $\text{OH"-("CH"_2)_2-"OH}$, so it has hydrogen-bonding intermolecular forces.

That means it can dissociate in water pretty easily in its current form (besides, its $\text{pKa}$ is close to that of water, so little of it would be deprotonated or protonated in water).

Therefore, for ethanediol, we can say that its van't Hoff factor is approximately $\textcolor{g r e e n}{i = 1}$.

Finally, we can get the solution freezing point from the equation we first listed:

$\Delta {T}_{f} = {T}_{f} - {T}_{f}^{\text{*}}$

= T_f - 0^@ "C" = (1.86^@ "C/m")("10.75 m")(1)

$\implies \textcolor{b l u e}{{T}_{f} \approx - {20}^{\circ} \text{C}}$

As I said before, it would be a freezing point depression, so it makes sense that ${T}_{f} < 0$.

So, there you go; the solution was indeed $\text{B}$, $\setminus m a t h b f \left({T}_{f} = - {20.1}^{\circ} \text{C}\right)$.