# What is the freezing point of a 1molal and 0.432molal Zn(H2O)_6 ^(+2)?

Mar 30, 2017

I'll do the 1 molal solution. You should then be able to do a 0.432 molal solution.

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

• ${T}_{f}$ is the freezing point, of course, and ${T}_{f}^{\text{*}}$ is that of water.
• $i$ is the number of ions in solution. We ignore ion pairing for simplicity.
• ${K}_{f} = {1.86}^{\circ} \text{C/m}$
• $m$ is the molality, traditionally in units of $\text{m}$ or "molal".

Clearly, water is not an ion, and the hexahydrate acts as the simple cation in water. Thus, $i = 1$, and we simply have:

$\Delta {T}_{f} = {T}_{f} - {0}^{\circ} \text{C} = \textcolor{b l u e}{{T}_{f}}$

$= - \left(1\right) \left({1.86}^{\circ} \text{C/m")("1 m") = color(blue)(-1.86^@ "C}\right)$