Chapter 13, Problem 127E

Interpretation Introduction

Interpretation:

The boiling point of an aqueous solution can be calculated with the help of freezing point temperature.

Concept introduction:

Molality is the number of moles of solute to the mass of solvent.

The mass of the solvent should be in kilograms $\left(\text{kg}\right)$.

Calculate molality as follows:

$m=\frac{\text{Moles solute}}{\text{Kilograms solvent}}$ …… (1)

The temperature at which change of liquid state to solid state takes is called freezing point.

The freezing point depression is as follows:

$\Delta T_{\text{f}}=m\times K_{\text{f}}$ …… (2)

Here, $\Delta T_{\text{f}}$ is the change in temperature of the freezing point in $\text{C}$, $m$ is the

molality of the solution in $\frac{\text{mol solute}}{\text{kg solvent}}$ and $K_{\text{f}}$ is the freezing point depression

constant for the solvent.

For water:

$K_{\text{f}}=1.86\frac{\text{C}\text{ kg solvent}}{\text{mol solute}}$

The boiling point is the temperature at which vapor pressure of the liquid becomes equal to the atmospheric pressure is known as boiling point.

The boiling point elevation is as follows:

$\Delta T_{\text{b}}=m\times K_{\text{b}}$ …… (3)

Here, $\Delta T_{\text{b}}$ is the change in temperature of the boiling point in $\text{C}$, $m$ is the

molality of the solution in $\frac{\text{mol solute}}{\text{kg solvent}}$ and $K_{\text{b}}$ is the boiling point elevation

constant for the solvent.

For water:

$K_{\text{b}}=0.512\frac{\text{C}\text{ kg solvent}}{\text{mol solute}}$