Chapter 13, Problem 62QP

Interpretation Introduction

Interpretation:

The boiling point and freezing point of solution of given molar concentration of naphthalene in benzene is to be calculated.

Concept introduction:

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

The freezing point depression is as follows:

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

Here, $\Delta T_{\text{f}}$ is the change in temperature of the freezing point in $\text{C}$, $m$ is the molality of the solution, and $K_{\text{f}}$ is the freezing point depression constant for the solvent.

For benzene

$K_{\text{f}}=5.12\frac{\text{C}}{m}$

The actual freezing point is the difference between freezing point of pure solvent and freezing point depression.

$T_f=T_f{}^{°}-\Delta T_{\text{f}}$

The temperature at which vapor pressure of a particular liquid has the same value as the atmospheric pressure is called its boiling point.

The boiling point elevation is as follows:

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

Here, $\Delta T_{\text{b}}$ is the change in temperature of the boiling point, $m$ is the molality of the solution, and $K_{\text{b}}$ is the boiling point elevation constant for the solvent.

For benzene:

$K_{\text{b}}=2.53\frac{\text{C}}{m}$

The actual boiling point of the solution is the sum of boiling point of pure solvent and the boiling point elevation $\left(\Delta T_{\text{b}}\right)$.

$T_b=T_b{}^{°}+\Delta T_{\text{b}}$