Chapter 14, Problem 100A

In the lab, you dissolve 179 g of $MgC{l}_2$ into1.00 L of water. Use Table 14.6 to find the freezing pointof the solution.

Interpretation Introduction

Interpretation:

The freezing point of a solution have to be calculated.

Concept introduction:

For electrolytic solutes, the depression of the freezing point is directly proportional to molalityof the solution.

Freezing point depression $\Delta {\mathrm{T}}_f=i\times K_f\times m$

$\Delta {\mathrm{T}}_f$ Depression of the freezing point = the freezing point difference between a solution and the pure solvent

$i$ is the van’t Hoff factor = for strong electrolyte $i$ is equal to number of dissociate ions.

$K_f$ is the molalfreezing point depression constant.

$m$ is the molality of the solution.

Answer to Problem 100A

The freezing point of the solution is − 10.5 °C.

Explanation of Solution

Given information:

Mass of MgCl₂ = 179 gm

Volume of water = 1000 mL

Write the expression for freezing point depression

$\Delta {\mathrm{T}}_f=i\times K_f\times m$

Write the expression for the number of moles of $MgC{l}_2$.

$\begin{array}{c}NumberofmolesofMgC{l}_2=\frac{MassofMgC{l}_2}{MolecularweightofMgC{l}_2}\\ =\frac{179gm}{95.2gm/mol}\\ =1.88mole\end{array}$

(Note: Molecular weight of MgCl₂ = 95.2 gm/mol)

Calculate the mass of water.

$\begin{array}{c}\mathrm{Mass}ofwater=Volumeofwater\times Densityofwater\\ =1000mL\times 1gm/mL\\ =1000gm\\ =1kg\end{array}$ \\

Calculate the molality of the solution.

$\begin{array}{c}MolalityofMgC{l}_{2}solution=\frac{Numberofmolesof MgC{l}_2}{ Mass of water(inKg)}\\ =\frac{1.88mole}{ 1kg}\\ =1.88m\end{array}$

MgCl₂is strong electrolyte, it dissociates completely in the solution to give three ions (one Mg²+ and two Cl- ion). So, i = 3

$\begin{array}{l}\Delta T_f=i\times K_f\times m\\ =3\times 1.86°C/m\times 1.88m\\ =10.5°C\end{array}$

The freezing point of the solution is:

$\begin{array}{c}{T}_f=0°\mathrm{C}–10.5°\mathrm{C}\\ =–10.5°\mathrm{C}\end{array}$

Thus, the freezing point of the solution is $T_f=–10.5°\mathrm{C}$.