Question

Calculate the freezing point of an aqueous solution containing $10.5{\text{ g}}$ of magnesium bromide in $200{\text{ g}}$ of water, assuming complete dissociation of magnesium bromide.
 (Molar mass of magnesium bromide $ = 184{\text{ g mo}}{{\text{l}}^{ - 1}}$, ${{\text{K}}_{\text{f}}}$ for water $ = 1.86{\text{ K kg mol}}{{\text{l}}^{ - 1}}$).

Answer 1

Hint: To solve this we must know that the freezing point of a pure solvent decreases when a non-volatile solute is added to it. This decrease in the freezing point is known as the depression in freezing point. Thus, when magnesium bromide is added to water the freezing point of water decreases.

Formulae Used:
1) $\Delta {T_f} = {K_f} \times m$
2) ${\text{Molality}}\left( {\text{m}} \right) = \dfrac{{{\text{Number of moles of solute}}\left( {{\text{mol}}} \right)}}{{{\text{Weight of solvent}}\left( {{\text{kg}}} \right)}}$
3) ${T_f} = {T_0} - \Delta {T_f}$

Complete step by step solution:
The equation for the depression in freezing point of a solution is,
$\Delta {T_f} = {K_f} \times m$
Where, $\Delta {T_f}$ is the depression in freezing point,
${K_f}$ is the constant of the depression in freezing point,
$m$ is the molality of the solution.

We know that molality is the number of moles of solute dissolved per kilogram of solvent. The mathematical expression for molality is as follows:
${\text{Molality}}\left( {\text{m}} \right) = \dfrac{{{\text{Number of moles of solute}}\left( {{\text{mol}}} \right)}}{{{\text{Weight of solvent}}\left( {{\text{kg}}} \right)}}$
We know that the number of moles is the ratio of mass to the molar mass. Thus,
${\text{Molality}}\left( {\text{m}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}} \times \dfrac{{\text{1}}}{{{\text{Weight of solvent}}\left( {{\text{kg}}} \right)}}$
Thus, the equation for the depression in freezing point of a solution is,
$\Delta {T_f} = {K_f} \times \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}} \times \dfrac{{\text{1}}}{{{\text{Weight of solvent}}\left( {{\text{kg}}} \right)}}$
Substitute $1.86{\text{ K kg mo}}{{\text{l}}^{ - 1}}$ for the ${{\text{K}}_{\text{f}}}$ of water, $10.5{\text{ g}}$ for the mass of solute i.e. magnesium bromide, $184{\text{ g mo}}{{\text{l}}^{ - 1}}$ for the molar mass of magnesium bromide, $200{\text{ g}} = 200 \times {10^{ - 3}}{\text{ kg}}$ for the mass of solvent i.e. water. Thus, $\Delta {T_f} = 1.86{\text{ K kg mo}}{{\text{l}}^{ - 1}} \times \dfrac{{10.5{\text{ g}}}}{{184{\text{ g mo}}{{\text{l}}^{ - 1}}}} \times \dfrac{{\text{1}}}{{200 \times {{10}^{ - 3}}{\text{ kg}}}}$
$\Delta {T_f} = 0.530{\text{ K}}$
Thus, the depression in freezing point is $0.530{\text{ K}}$.
Now, calculate the freezing point using the equation as follows:
${T_f} = {T_0} - \Delta {T_f}$
Where, ${T_f}$ is the freezing point,
${T_0}$ is the temperature at ${0^ \circ }{\text{C}}$
$\Delta {T_f}$ is the depression in freezing point.
Substitute $273{\text{ K}}$ for the temperature at ${0^ \circ }{\text{C}}$, $0.530{\text{ K}}$ for the depression in freezing point.
Thus,
${T_f} = \left( {273 - 0.530} \right){\text{ K}}$
${T_f} = 272.47{\text{ K}}$

Thus, the freezing point of an aqueous solution containing $10.5{\text{ g}}$ of magnesium bromide in $200{\text{ g}}$ of water is $272.47{\text{ K}}$.

Note: The freezing point of a pure solvent decreases when any non-volatile solute is added to it. This is known as depression in freezing point. The freezing point of a pure solvent is always higher than that of its solution.This is one of the colligative properties ,those properties which depend upon the number of molecules present in the solution.