# What is the freezing point for an aqueous solution with "0.195 molal" "K"_2"S" dissolved in it? Assume 100% dissociation. K_f = 1.86^@ "C/m".

Aug 9, 2017

As you should have seen in your book,

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

where:

• $\Delta {T}_{f}$ is the change in freezing point in $\text{^@ "C}$, from that of the pure solvent, ${T}_{f}^{\text{*}}$, to that of the solution, ${T}_{f}$.
• $i$ is the van't Hoff factor, i.e. the effective number of solute particles in solution.
• ${K}_{f} = {1.86}^{\circ} \text{C/m}$ is the freezing point depression constant of water. The negative sign is in the equation.
• $m$ is the molality of the solution... $\text{mol solute/kg solvent}$. What is the solute?

Assuming 100% dissociation...

${\text{K"_2"S"(aq) -> 2"K"^(+)(aq) + "S}}^{2 -} \left(a q\right)$

and $1 + 2 = 3 \approx i$, so...

$\textcolor{b l u e}{{T}_{f}} = {T}_{f}^{\text{*}} - i {K}_{f} m$

$= {0}^{\circ} \text{C" - 3 cdot 1.86^@ "C/m" cdot "0.195 m}$

$= \textcolor{b l u e}{\underline{- {1.088}^{\circ} \text{C}}}$

What was the change in freezing point?