Chapter 16, Problem 88A

Interpretation Introduction

Interpretation: The boiling point of a solution is to be calculated from its given freezing point $-2.47°\text{C}$ .

Concept introduction: The addition of solute particles to a particular solvent will decrease the freezing point and raise the boiling point of that solvent.

The formula used for calculating the depression in the freezing point is,

  $\Delta T_f=T_{Solvent}-T_{Solution}$

Here, $\Delta T_f$ is the freezing point depression, $T_{Solvent}$ is the freezing point of the solvent, and $T_{Solution}$ is the freezing point of the solution.

The formula used for calculating the elevation in boiling point is,

  $\Delta T_b=K_{b}m$

Here, $\Delta T_b$ is the elevation in boiling point, $K_b$ is the molal boiling point elevation constant, and $m$ is molality.

Answer to Problem 88A

The boiling point of the given solution is $102.47°\text{C}$ .

Explanation of Solution

The freezing point of the solution is given as $T_{Solution}=-2.47°\text{C}$ .

The freezing point of the solvent water is given as $0°\text{C}$ .

The boiling point of the solvent, water is $100°\text{C}$ .

Substitute $-2.47°\text{C}$ for $T_{Solution}$ and $0°\text{C}$ for $T_{Solvent}$ in the equation of freezing point depression.

  $\begin{array}{c}\Delta T_f=0°\text{C}-\left(-2.47°\text{C}\right)\\ =2.47°\text{C}\end{array}$

At equilibrium, the value of $K_b$ and $K_f$ are equal. And, also molality of the solution is equal.

Therefore,

  $\frac{\Delta T_b}{\Delta T_f}=1$

Hence,

  $\begin{array}{c}\Delta T_b=\Delta T_f\\ =2.47°\text{C}\end{array}$

Substitute $2.47°\text{C}$ for $\Delta T_f$ and $100°\text{C}$ for $T_{Solvent}$ in the equation of freezing point depression.

  $\begin{array}{c}2.47°\text{C}=100°\text{C}-T_{Solution}\\ T_{Solution}=102.47°\text{C}\end{array}$

The boiling point of the given solution is $102.47°\text{C}$ .

Conclusion

So, the boiling point of the given solution is $102.47°\text{C}$ .