# What is the molar mass of the solute if "4.18 g" of it dissolved in "36.30 g" of benzene (K_f = 5.12^@ "C"cdot"kg/mol") generates a solution with a freezing point of 2.70^@ "C"? The freezing point of benzene is 5.53^@ "C".

Feb 16, 2017

$\text{208.33 g/mol}$

Once you recognize this is freezing point depression, you can use the following equation:

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

where:

• ${T}_{f}$ and ${T}_{f}^{\text{*}}$ are the freezing points of the solution and pure solvent, respectively.
• $i$ is the van't Hoff factor. For non-ionic solutes, i.e. nonelectrolytes, $i = 1$, as there is only one "dissociated" particle per dissolved particle.
• ${K}_{f} = {5.12}^{\circ} \text{C"cdot"kg/mol}$ is the freezing point depression constant for benzene, the solvent.
• $m$ is the molality of the solution, i.e. $\text{mols solute"/"kg solvent}$.

As mentioned, benzene is the solvent, because its ${K}_{f}$ is given, and ${K}_{f}$ is only for solvents. Therefore, we should calculate the mols of the solute and kg of benzene.

Simply from the units of molality, we have:

$m = \left(\text{solute mass" xx 1/"molar mass")/("kg solvent}\right)$

$= \left(\text{4.18 g"xx"mol"/"g solute")/("0.03630 kg benzene}\right)$

Therefore, we can solve for the molar mass later, as long as we can calculate the molality.

DeltaT_f = 2.70^@ "C" - 5.53^@ "C" = -(1)(5.12^@ "C"cdot"kg/mol")(m)

$\implies m = \text{0.5527 mols solute"/"kg solvent}$

Now, we can solve for the molar mass.

$\textcolor{b l u e}{\text{Molar mass") = ("4.18 g solute")/("0.5527 mols solute"/cancel"kg solvent" xx 0.03630 cancel"kg solvent}}$

$=$ $\textcolor{b l u e}{\text{208.33 g/mol}}$