Chapter 6, Problem 6.76QP

Interpretation Introduction

Interpretation:

The boiling temperature of $0.50mNaCl$ and sucrose solution 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 6.76QP

The boiling temperature of $0.50mNaCl$ and sucrose solution is $100.52^{o}C$ and $100.26^{o}C$ respectively.

Explanation of Solution

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

$NaCl$ being ionic compound, dissociate to form two particles.  So $0.50mNaCl$ is $1m\left(0.5m\times 2\right)$ particles.

So boiling temperature of $0.50mNaCl$  can be calculated as follows,

    $\begin{array}{c}\Delta T_b=k_b\left(mparticles\right)\\ \\ =\left(\frac{{0.52}^{o}C}{m}\right)\left(1m\right)\\ \\ =0.52{}^{o}C\\ \\ T_1-T_2=0.52{}^{o}C\end{array}$

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.

    $\begin{array}{c}{T}_1-T_2=0.52{}^{o}C\\ \\ T_1-{100}^{o}C=0.52{}^{o}C\\ \\ T_1=0.52{}^{o}C+{100}^{o}C\\ \\ T_1=100.52^{o}C.\end{array}$

The boiling temperature of sodium chloride solution is $100.52^{o}C.$

Boiling temperature of $0.50m$ sucrose can be calculated as follows,

    $\begin{array}{c}\Delta T_b=k_b\left(mparticles\right)\\ \\ =\left(\frac{{0.52}^{o}C}{m}\right)\left(0.5m\right)\\ \\ =0.26{}^{o}C\\ \\ T_1-T_2=0.26{}^{o}C\end{array}$

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.

    $\begin{array}{c}{T}_1-T_2=0.26{}^{o}C\\ \\ T_1-{100}^{o}C=0.26{}^{o}C\\ \\ T_1=0.26{}^{o}C+{100}^{o}C\\ \\ =100.26^{o}C.\end{array}$

Boiling temperature of sucrose solution is $100.26^{o}C.$