Solutions
1.0 Solution
2.0 Methods of expressing concentration of a Solution
3.0 Solubility
4.0 Henry's law
5.0 Raoult's law
6.0 Azotropes
7.0 Colligative Properties
7.1 Relative lowering of vapour pressure
7.2 Elevation in Boiling point
7.3 Depression of Freezing point
7.4 Osmosis and Osmotic pressure
8.0 Abnormal Colligative Properties
7.2 Elevation in Boiling point
7.2 Elevation in Boiling point
7.3 Depression of Freezing point
7.4 Osmosis and Osmotic pressure
Boiling point is temperature at which vapour pressure of liquid becomes equal to pressure over the lid of container (if pressure over liquid equals atmospheric pressure then boiling point will be called Normal boiling point).
So, if a liquid have high vapour pressure then it means its boiling point is low and vice versa. Basically, vapour pressure and boiling point have inverse relation.
As from last section, vapour pressure decreases on addition of non volatile solute. So, it implies boiling point increases with addition of non volatile solute.
For example: Vapour pressure of solution of sucrose in water is less then atmospheric pressure at $373.15\ K$ (boiling point of water). So, boiling point of solution is always high than solvent.
Elevation in boiling point depends on number of solute particles.
Change in boiling point is given by $\Delta {T_b}$ and
$$ \Delta {T_b} = {T_{solution}} - {T_{solvent}} $$
Experimentally it is found that $\Delta {T_b}$ is directly proportional to molal concentration of solute.
$$\begin{equation} \begin{aligned} \Delta {T_b} \propto m \\ \Delta {T_b} = {K_b}m \\\end{aligned} \end{equation} $$
where ${K_b}$ is called molal elevation constant or Ebullioscopic constant.
Unit of ${K_b}$ is $Kkgmo{l^{ - 1}}$
$$m = ({w_2} \times 1000)/{w_1} \times {M_2}$$
$$ \Rightarrow \Delta {T_b} = {K_b} \times ({w_2} \times 1000)/{w_1} \times {M_2}$$
From this realtion molecular mass of solute(${M_2}$) can be calculated.
Also,
$${K_b} = RT_b^2M/\Delta {H_v}$$
where, $M$ is molar mass of solvent and $\Delta {H_v}$ is Heat of vapourization per mole of solvent.
Relation between Relative lowering of vapour pressure and Elevation in boiling point is as follows:
As,
$$ \Rightarrow \Delta {T_b} = {K_b} \times ({w_2} \times 1000)/{w_1} \times {M_2}$$
and
$$ \Rightarrow \Delta {P_1}/P_1^0 = {n_2}/{n_1} = {w_2}{M_1}/{M_2}{w_1}$$
using both we can easily get,
$$\Delta {T_b} = {K_b} \times 1000((P_A^0 - {P_A})/P_A^0 \times {M_{solvent}}$$
