# If k_f=1.86*""^(@)C*mol^-1*kg for water, what is the freezing point of a solution composed of 255*kg CaCl_2, and 33.8*L of water?

Apr 27, 2017

Well, $\Delta {T}_{\text{fusion}} = {k}_{f} \times c \times i$ where.................

#### Explanation:

Where ${k}_{f}$ is the $\text{molal freezing point depression constant}$, which you have kindly given, $c = \text{molal concentration}$ in $m o l \cdot k {g}^{-} 1$, and $i$, the so-called $\text{van't Hoff factor}$ which reflects the speciation of an ionic solid in a solvent.

Clearly, here, $i = 3$. Why? Because in solution, $\text{calcium chloride}$ speciates to give three ions......

$C a C {l}_{2} \left(a q\right) \stackrel{{H}_{2} O}{\rightarrow} C {a}^{2 +} + 2 C {l}^{-}$

And so...............

${\Delta}_{\text{fusion"=1.86*}}^{\circ} C \cdot {m}^{-} 1 \times 3 \times \frac{\frac{2.55 \times {10}^{3} \cdot g}{110.98 \cdot g \cdot m o {l}^{-} 1}}{33.8 \cdot k g} = 3.80$ ""^@C.

This reflects the depression of the fusion point (the melting point) with RESPECT to PURE SOLVENT. And thus $\text{fusion point}$ $\equiv$ $- 3.80$ ""^@C.

In practice, the melting point probably would not be depressed so far. And this is the province of measurement.