# Question 8ad06

Jan 10, 2015

The answer is $6.33$ $\text{g}$ of $M n {I}_{2}$ must be used to get that drop in freezing point temperature.

The mathematical equation for boiling point depression, which is a colligative property, is this

$\Delta {T}_{f} = {K}_{f} \cdot m \cdot i$, where

$\Delta {T}_{f}$ - the freezing point depression, defined as the difference between the freezing temperature of the pure solvent minus the freezing temperature of the solution, ${T}_{f} \left(\text{pure [solvent](http://socratic.org/chemistry/solutions-and-their-behavior/solvent)") - T_f("solution}\right)$;

${K}_{f}$ - the molal freezing point depression constant of the solvent -${1.86}^{\circ} C \cdot \text{kg/mol}$ for water;

$m$- the molality of the solution;
$i$ - the number of dissolved particles (Van't Hoff Factor).

You must use the equation above to determine the molality of the solution, i.e. the moles of $M n {I}_{2}$ you have in that amount of water. The moles of $M n {I}_{2}$ will then get you the mass you need to determine. So, you know that $\Delta {T}_{f}$ is equal to ${0.4500}^{\circ} C$.

Since $M n {I}_{2}$ is soluble in water, it will dissociate into $M {n}^{2 +}$ and $2 {I}^{-}$ ions, which will give you a Van't Hoff factor of 1 + 2 = 3. Every molecule of $M n {I}_{2}$ creates 3 ions when placed in water, 1 $M {n}^{2 +}$ cation and 2 ${I}^{-}$ anions. So,

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

$m = \frac{{0.4500}^{\circ} C}{{1.86}^{\circ} C \cdot \frac{k g}{m o l} \cdot 3} = 0.0806$ $\text{molal}$

You know that molality is defined as moles of solute divided by mass of solution (in kg). This will get you

m = n_("solute")/m_("solvent") = n_(MnI_2)/(m_("water")) => n_(MnI_2) = m * 254.0 * 10^(-3)kg#

${n}_{M n {I}_{2}} = 0.0806 \cdot 254.0 \cdot {10}^{- 3}$ $\text{kg} = 0.0205$ $\text{moles}$

This means that the mass of $M n {I}_{2}$ needed is

${m}_{M n {I}_{2}} = {n}_{M n {I}_{2}} \cdot \text{molar mass}$

${m}_{M n {I}_{2}} = 0.0205$ $\text{moles} \cdot 308.7$ $\text{g/mol} = 6.33$ $\text{g}$