# Question f0370

Jan 29, 2017

${T}_{f} = - {1.58}^{\circ} \text{C}$

This should be in your textbook under the topic of "Colligative Properties". Basically, any solute you put into a solvent decreases its freezing point and increases its boiling point. This is called a colligative property of the solute. We call these changes in phase transition temperatures the freezing point depression and boiling point elevation, respectively.

Here, the solute is ethylene glycol and the solvent is water. Again, it does not matter what the solute is, unless you need to know how many $\text{mol}$s you have and you were given its mass in $\text{g}$.

The change in freezing point is given by:

$\boldsymbol{\Delta {T}_{f} = {T}_{f} - {T}_{f}^{\text{*}} = - i {K}_{f} m}$,

where:

• ${T}_{f}$ is the freezing point of the solution (the solvent plus solute).
• ${T}_{f}^{\text{*}}$ is the freezing point of the pure solvent (without solute).
• $i$ is the van't Hoff factor. It is approximately the number of particles in solution that are made for each particle of the solute that is placed into solution. Therefore, for nonelectrolytes, $i = 1$.
• ${K}_{f}$ is the freezing point depression constant. For water, ${K}_{f} = \text{1.86"^@ "C/m}$, or ${1.86}^{\circ} \text{C"cdot"kg/mol}$.
• $m$ is the molality, the molal concentration of the solution in $\text{mols solute"/"kg solvent}$. This is not to be confused with molarity, $\text{mols solute"/"L solution}$.

First, get the change in freezing point. It's OK if you get a sign error and the number is not negative, as long as you know it should be negative.

DeltaT_f = -iK_fm = -(1)(1.86^@ "C"cdotcancel"kg/mol")(0.85 cancel("mols solute"/"kg solvent"))

$= - {1.58}^{\circ} \text{C}$

Now you can calculate the new freezing point, which is lower than that of the pure solvent.

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

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

= 0.00^@ "C" + (-1.58^@ "C")#

$\implies \textcolor{b l u e}{{T}_{f} = - {1.58}^{\circ} \text{C}}$