Chapter 14, Problem 99A

Interpretation Introduction

Interpretation:

The freezing point of a solution have to be calculated.

Concept introduction:

For nonvolatile and nonelectrolyte solutes, the depression of the freezing point is directly proportional tomolalityof the solution.

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

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

$K_f$ is the molalfreezing point depression constant.

$m$ is the molality of the solution.

Answer to Problem 99A

The freezing point of the solution is 2.74 °C.

Explanation of Solution

Write the expression for Molality of the solution.

$\mathrm{Molality}ofthesolution=\frac{Numberofmolesof the solute}{ Mass of the solvent(inKg)}$

$\begin{array}{l}Numberofmolesofnaphthalene=\frac{Massofnapthalene}{Molecularweightofnapthalene}\\ =\frac{12.1gm}{128.08gm/mol}=0.0945mole\end{array}$

(Note: Molecular weight of naphthalene = 128.08 gm/mol)

$\begin{array}{l}Molalityofthesolution=\frac{Numberofmolesof naphthalene}{ Mass of benzene(inKg)}\\ =\frac{0.0945mole}{ 0.175kg}=0.54m\end{array}$

Write the expression for freezing point depression

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

Substitute the values in the above expression.

$\begin{array}{l}\Delta T_f=K_f\times m\\ =5.12°C/m\times 0.54m\\ =2.76°C\end{array}$

The freezing point of the solution:

$\begin{array}{c}{T}_f=5.5°\mathrm{C}-2.76°\mathrm{C}\\ =2.74°\mathrm{C}\end{array}$

Thus, the freezing point of the solution is $2.74°\mathrm{C}$.