# What is the molality for an aqueous solution whose freezing point dropped by 15^@ "C" due to addition of "CaCl"_2(s)? K_f = 1.86^@ "C/m" for water.

Nov 17, 2016

I got $\text{2.69 m}$.

In general, the decrease in freezing point can be found from the following equation:

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

$= \boldsymbol{i {K}_{f} m}$

where:

• ${T}_{f}$ is the freezing point of the ${\text{CaCl}}_{2} \left(a q\right)$ solution.
• $\text{*}$ means of pure water. That is, ${T}_{f}^{\text{*" = 0^@ "C}}$.
• $i$ is the van't Hoff factor. A good way to estimate it is to say that the number of ions the solute makes is equal to $i$.
• ${K}_{f} = - {1.86}^{\circ} \text{C/m}$ is the freezing point depression constant of water near ${0}^{\circ} \text{C}$. The units are read as degrees celcius per molal.
• $m = \text{mol solute"/"kg solvent}$ is the molality of the solution. The unit is read molal.

To solve for the molality:

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

Now, we would estimate that $\boldsymbol{i \approx 3}$ since ${\text{CaCl}}_{2} \left(a q\right)$ dissociates as follows:

${\text{CaCl"_2(s) stackrel("H"_2"O"(l))(->) "Ca"^(2+)(aq) + 2"Cl}}^{-} \left(a q\right)$

giving $\boldsymbol{3}$ ions in solution. Finally, we know that ${T}_{f}^{\text{*" = 0^@ "C}}$. So plugging the numbers in gives:

$\textcolor{b l u e}{m} = \left(- {15}^{\circ} \cancel{\text{C" - 0^@ cancel"C")/(3*(-1.86^@ cancel"C""/m}}\right)$

$=$ $\textcolor{b l u e}{\text{2.69 m}}$ (molals)