Chapter 13, Problem 128E

Interpretation Introduction

Interpretation:

The freezing point of an aqueous solution that boils at $102.1{}^{\circ }\text{C}$ is to be calculated.

Concept Introduction:

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

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

Formula of molality is:

$m=\frac{\text{Moles solute}}{\text{Kilograms solvent}}$

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

The freezing point depression is:

$\Delta T_{\text{f}}=m\times K_{\text{f}}$

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.

The boiling point elevation is:

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

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