7.3 Depression of Freezing point

2.1 $n$-factor

3.1 Solubility of gas in liquids

5.1 Vapour pressure of solids in liquid
5.2 Ideal Solution
5.3 Non Ideal Solution

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

The temperature at which vapour pressure of the substance in its liquid phase is equal to its vapour pressure in solid phase.

The lowering of vapour pressure of solution leads to lowering of freezing point compared to pure solvent.

Basically, a solution freezes when its vapour pressure equals vapour pressure of solid solvent.

As clear from the graph,

If,

$T_f^0$ = Freezing point of pure solvent

${T_f}$ = Freezing point of solution ( containing nonvolatile solute)

From graph,

$$T_f^0 > {T_f}$$

then, depression in freezing point is $$\Delta {T_f} = T_f^0 - {T_f}$$

Experimentally,

$$\Delta {T_f} \propto m$$

$$\Delta {T_f} = {K_f}m$$

where,

$m$ is molal concentration of solute in solution and ${K_f}$ is called freezing point depression constant or molal depression constant or cryoscopic constant.

If ${w_2}$ gram of solute having ${M_2}$ molar mass is dissolved in ${w_1}$ gram of solvent, then

$$m = \frac{{{w_2} \times 1000}}{{{M_2} \times {w_1}}}$$

$$ \Rightarrow \Delta {T_f} = {K_f} \times \left( {\frac{{{w_2} \times 1000}}{{{M_2} \times {w_1}}}} \right)$$

By knowing all unknowns, we can calculate the molar mass of solute.

Also,

$${K_f} = \frac{{R \times {M_1} \times T_f^2}}{{1000 \times \Delta {H_{fus}}}}$$

Sometimes, this relation is also required in problems.