# A solution that contains "13.2 g" of solute in "250 g" of "CCl"_4 freezes at -33^@ "C". What is the FW of the solute? T_f^"*" of "CCl"_4 is -22.8^@ "C" and its K_f = -29.8^@ "C/m".

##### 1 Answer
Aug 12, 2017

$M = \text{154.26 g/mol}$

You can think of this as just an extension of freezing point depression into solvents that aren't water. Thus, refer to the freezing point depression equation:

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

where:

• ${T}_{f} = - {33}^{\circ} \text{C}$ is the freezing point, and ${T}_{f}^{\text{*" = -22.8^@ "C}}$ is for the pure solvent.
• $\Delta {T}_{f}$ is the change in freezing point, which will end up being negative.
• $i$ is the van't Hoff factor, i.e. the effective number of solute particles in solution after any potential dissociation.
• ${K}_{f} = {29.8}^{\circ} \text{C/m}$ is the freezing point depression constant of ${\text{CCl}}_{4}$. (Yours is negative, but I put the negative in the equation instead.)
• $m$ is the molality of the solution, in $\text{mols solute/kg solvent}$.

Now, the question hasn't specified whether the solute is a nonelectrolyte. However, since ${\text{CCl}}_{4}$ is nonpolar, it dissolves a solute that is also nonpolar, which is not likely to dissociate in solution, so $i = 1$.

The molality is what we can end up using to get the formula mass ("FW") of the solute, so we would solve for the molality.

$m = - \frac{\Delta {T}_{f}}{i {K}_{f}}$

$= - \left(- {33}^{\circ} \text{C" - (-22.8^@ "C"))/((1) cdot 29.8^@ "C/m}\right)$

$= 0.342$ $\text{mols/kg}$

Knowing that we have $\text{250 g}$ of the solvent, or $\text{0.250 kg}$, we therefore have:

$\text{0.342 mols"/"kg" = "0.0856 mols"/"0.250 kg}$

As a result, we have $\text{0.0856 mols}$ of solute for every $\text{13.2 g}$ of solute. Therefore, the formula mass is given by:

color(blue)(M) = "13.2 g"/("0.0856 mol") = color(blue)("154.26 g/mol")