Chapter 6, Problem 6.73QP

(a)

Interpretation Introduction

Interpretation:

The freezing temperature of $1.50m$ urea has to be calculated.

Concept Introduction:

Depression in freezing point:

At freezing point solid phase of the solvent will be in equilibrium with liquid phase.  When solute molecule is added, it will interfere with the rate at which the liquid molecules associate to form solid state, hence lower the freezing point of the solution.

    $\Delta T_f=k_f\left(mparticles\right)$

Where, $\Delta T_f$ difference in temperature of pure solvent and solution, $k_f$ is freezing point depression constant.

Answer to Problem 6.73QP

The freezing temperature of $1.50m$ urea solution is $-2.79{}^{o}C.$

Explanation of Solution

$k_f$ for aqueous solution is $\frac{{1.86}^{o}C}{m}.$

So freezing point temperature of $1.50m$ urea can be calculated as follows,

    $\begin{array}{c}\Delta T_f=k_f\left(mparticles\right)\\ \\ =\left(\frac{{1.86}^{o}C}{m}\right)\left(1.5m\right)\\ \\ =2.79{}^{o}C\\ \\ T_1-T_2=2.79{}^{o}C\end{array}$

Where $T_1$ is the freezing point of pure water and $T_2$ is the freezing point of the urea solution.

Therefore freezing point of the solution is given below.

    $\begin{array}{c}{T}_1-T_2=2.79{}^{o}C\\ \\ 0{}^{o}C-T_2=2.79{}^{o}C\\ \\ T_2=0{}^{o}C-2.79{}^{o}C\\ \\ =-2.79{}^{o}C.\end{array}$

Freezing temperature of the solution is $-2.79{}^{o}C.$

(b)

Interpretation Introduction

Interpretation:

The freezing temperature of $1.50mLiBr$ has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 6.73QP

The freezing temperature of $1.50mLiBr$ is $-5.58{}^{o}C.$

Explanation of Solution

$k_f$ for aqueous solution is $\frac{{1.86}^{o}C}{m}.$

$LiBr$ being ionic compound, dissociate to form two particles.  So $1.50mLiBr$ is $3m\left(1.5m\times 2\right)$ particles.

So freezing point temperature of $1.50mLiBr$  can be calculated as follows,

    $\begin{array}{c}\Delta T_f=k_f\left(mparticles\right)\\ \\ =\left(\frac{{1.86}^{o}C}{m}\right)\left(3m\right)\\ \\ =5.58{}^{o}C\\ \\ T_1-T_2=5.58{}^{o}C\end{array}$

Where $T_1$ is the freezing point of pure water and $T_2$ is the freezing point of the solution.

Therefore freezing point of the solution is given below.

    $\begin{array}{c}{T}_1-T_2=5.58{}^{o}C\\ \\ 0{}^{o}C-T_2=5.58{}^{o}C\\ \\ T_2=0{}^{o}C-5.58{}^{o}C\\ \\ =-5.58{}^{o}C.\end{array}$

Freezing point temperature of the solution is $-5.58{}^{o}C.$