# Assume that you have a solution of an unknown solute in cyclohexane. If the solution has a freezing-point depression of 9.50Celcius, what is the molality of this solution? (The molal freezing-point constant of cyclohexane is 20.2 C/m)

Jun 28, 2017

We obtain a molality dependent on the assumed van't Hoff factor of $i \approx 1$,

$m \approx 0.470$ $\text{molal}$

We refer to the freezing point depression given by

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

where:

• ${T}_{f}$ is the freezing point in $\text{^@ "C}$ of the solution, and $\text{*}$ indicates pure solvent.
• $i$ is the van't Hoff factor of the solute, the effective number of dissociated particles per formula unit. For nonelectrolytes, this would be $1$, but we have no idea what this solute is...
• ${K}_{f} = {20.2}^{\circ} \text{C/m}$ is the freezing point depression constant of cyclohexane at its normal freezing point.
• $m$ is the molality in $\text{mol solute/kg solvent}$.

The molality expression is therefore

$\textcolor{b l u e}{m} = - \frac{\Delta {T}_{f}}{i {K}_{f}} = - \frac{1}{i} \left(- {9.50}^{\circ} \text{C")/(20.2^@ "C/m}\right)$

$= \textcolor{b l u e}{\frac{0.470}{i}}$ $\textcolor{b l u e}{\text{mol solute/kg solvent}}$

So we will have to provide a van't Hoff factor to determine the molality here.

Since this is a fairly high molality (high being higher than $\text{0.01 molal}$), we have to assume the solute is quite soluble, and thus that it is nonpolar and organic, making it a nonelectrolyte.

That means $i \approx 1$.

Of course, had the change in temperature been low enough that $m < \text{0.01 molal}$, we would have been in the dark about what kind of solute this could have been, as many things are barely soluble in cyclohexane...