Depression in Freezing Point | Colligative Properties Class 12

Depression in Freezing Point

Cryoscopy | Colligative Properties | Solutions Class 12

1. What is Freezing Point Depression?

The freezing point is the temperature at which the solid and liquid phases of a substance have the same vapor pressure. When a non-volatile solute is added to a solvent, the vapor pressure of the liquid decreases.

                    Concept: Because the solution's vapor pressure is lower than the pure solvent's, the solution curve intersects the solid solvent curve at a lower temperature. This decrease is called Depression in Freezing Point.
                

2. Mathematical Expression

If $T_f^\circ$ is the freezing point of pure solvent and $T_f$ is the freezing point of the solution:

$$ \Delta T_f = T_f^\circ - T_f $$

Experimentally, for dilute solutions, the depression ($\Delta T_f$) is directly proportional to the molal concentration ($m$) of the solute.

$$ \Delta T_f = K_f \times m $$

Where:

$m$ = Molality (moles of solute / kg of solvent)
$K_f$ = Molal Depression Constant or Cryoscopic Constant.

Unit of $K_f$: $K \cdot kg \cdot mol^{-1}$.

3. Calculation of Molar Mass ($M_2$)

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

$$ M_2 = \frac{1000 \times K_f \times w_2}{w_1 \times \Delta T_f} $$

Where $w_2$ is mass of solute and $w_1$ is mass of solvent (in g).

4. Thermodynamic Relation for $K_f$

$K_f$ depends only on the nature of the solvent (e.g., Water $K_f = 1.86$).

$$ K_f = \frac{R \cdot M_1 \cdot (T_f^\circ)^2}{1000 \cdot \Delta_{fus}H} $$

Where:

$R$ = Gas Constant
$M_1$ = Molar mass of solvent
$\Delta_{fus}H$ = Enthalpy of Fusion

5. Applications

Anti-freeze: Ethylene glycol is added to car radiators to lower the freezing point of water, preventing the engine from freezing in cold climates.
De-icing Roads: $NaCl$ or $CaCl_2$ is sprinkled on icy roads. They depress the freezing point of water, causing ice to melt at sub-zero temperatures.

6. For Electrolytes (Van't Hoff Factor)

If the solute undergoes association or dissociation, we include the factor $i$.

$$ \Delta T_f = i \times K_f \times m $$

Example: $CaCl_2$ ($i=3$) depresses freezing point 3 times more than Urea ($i=1$) for the same molality.

Practice Quiz

Test your knowledge on Cryoscopy.

Search This Blog

Translate