Chapter 6, Problem 6.74QP

(a)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Elevation in boiling point:

Boiling point is the temperature at which vapor pressure becomes equal to atmospheric pressure.  If a non volatile solute is added then the vapor pressure get lowered.  So, more temperature has to be provided for vaporizing.  Hence the boiling point increases.

    $\Delta T_b=k_b\left(mparticles\right)$

Where, $\Delta T_b$ difference in temperature of pure solvent and solution, $k_b$ is boiling point elevation constant.

Answer to Problem 6.74QP

The boiling temperature of $1.50m$ urea solution is $100.78^{o}C.$

Explanation of Solution

$k_b$ for aqueous solution is $\frac{{0.52}^{o}C}{m}.$

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

    $\begin{array}{c}\Delta T_b=k_b\left(mparticles\right)\\ \\ =\left(\frac{{0.52}^{o}C}{m}\right)\left(1.5m\right)\\ \\ =0.78{}^{o}C\\ \\ T_1-T_2=0.78{}^{o}C\end{array}$

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

Therefore boiling point of the solution is given below.

    $\begin{array}{c}{T}_1-T_2=0.78{}^{o}C\\ \\ T_1-{100}^{o}C=0.78{}^{o}C\\ \\ T_1=0.78{}^{o}C+{100}^{o}C\\ \\ =100.78^{o}C.\end{array}$

Boiling temperature of the solution is $100.78^{o}C.$

(b)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

Answer to Problem 6.74QP

The boiling temperature of $1.50mLiBr$ is $101.56^{o}C.$

Explanation of Solution

$k_b$ for aqueous solution is $\frac{{0.52}^{o}C}{m}.$

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

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

    $\begin{array}{c}\Delta T_b=k_b\left(mparticles\right)\\ \\ =\left(\frac{{0.52}^{o}C}{m}\right)\left(3m\right)\\ \\ =1.56{}^{o}C\\ \\ T_1-T_2=1.56{}^{o}C\end{array}$

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

Therefore boiling point of the solution is given below.

    $\begin{array}{c}{T}_1-T_2=1.56{}^{o}C\\ \\ T_1-{100}^{o}C=1.56{}^{o}C\\ \\ T_1=1.56{}^{o}C+{100}^{o}C\\ \\ T_1=101.56^{o}C.\end{array}$

The boiling temperature of the solution is $101.56^{o}C.$