Chapter 13.6, Problem 1PPA

Interpretation Introduction

Interpretation:

The freezing point and the boiling point of the solution are to be calculated with the help of mass of ethylene glycol and mass of water.

Concept introduction:

Molality is defined as the total number of moles present in one kilogram of the solvent.

The mass of the solvent should be in kilograms $\left(\text{kg}\right)$. Mathematically, molality can be expressed as follows:

$\text{Molality }\left(m\right)=\frac{\text{moles of solute}}{\text{mass of solvent }\left(\text{in Kg}\right)}$

$\text{No}\text{. of moles}=\frac{\text{mass}}{\text{molar mass}}$

The temperature at which the change from 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 °C, $m$ is the molality of the solution, and $K_{\text{f}}$ is the freezing point depression constant for the solvent.

For water:

$K_{\text{f}}=1.86\frac{\text{C}}{m}$

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 the liquid becomes equal to the atmospheric pressure is known as boiling point.

The boiling point elevation is as follows:

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

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 water:

$K_{\text{b}}=0.52\frac{\text{C}}{m}$

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

$T_b=T_b{}^{°}+\Delta T_{\text{b}}$