Freezing Point Depression in Solutions

Freezing Point Depression

Introduction to Freezing Point Depression

When a solute is added to a pure solvent, the freezing point of the solvent decreases. This phenomenon is known as freezing point depression, a type of colligative property that depends on the number of solute particles in a solution, not their identity.

• Normal Freezing Point: The temperature at which a pure solvent solidifies under standard atmospheric pressure.

• Freezing Point of Solution: The temperature at which a solution solidifies, which is lower than that of the pure solvent.

Key Terms and Definitions

• Colligative Properties: Properties that depend on the concentration of solute particles, not their chemical identity (e.g., boiling point elevation, freezing point depression, osmotic pressure).

• Freezing Point Depression (ΔTf): The decrease in the freezing point of a solvent upon addition of a solute.

• Molality (m): The number of moles of solute per kilogram of solvent.

• Cryoscopic Constant (Kf): A proportionality constant specific to each solvent, representing the freezing point depression per molal concentration of a non-volatile solute.

• Van't Hoff Factor (i): The number of particles into which a solute dissociates in solution. For non-electrolytes, i = 1.

Freezing Point Depression Formula

The freezing point depression can be calculated using the following equation:

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

• $\Delta T_f$ = change in freezing point (°C)

• $i$ = van't Hoff factor (number of particles per formula unit)

• $K_f$ = freezing point depression constant (°C·kg/mol)

• $m$ = molality (mol solute/kg solvent)

Constants for Water

Solvent	Normal Freezing Point (°C)	Kf (°C·kg/mol)
Water	0.00	1.86
Chloroform	-63.5	4.68
Benzene	5.5	5.12

Example Calculation

Example: Calculate the freezing point of a solution containing 110.7 g glucose (C6H12O6) dissolved in 302.6 g water.

• Moles of glucose: $\frac{110.7\ \text{g}}{180.16\ \text{g/mol}} = 0.6147\ \text{mol}$

• Mass of water: 302.6 g = 0.3026 kg

• Molality: $m = \frac{0.6147\ \text{mol}}{0.3026\ \text{kg}} = 2.03\ \text{m}$

• For glucose, $i = 1$ (non-electrolyte)

• $\Delta T_f = 1 \times 1.86 \times 2.03 = 3.78^\circ\text{C}$

• Freezing point of solution: $0.00^\circ\text{C} - 3.78^\circ\text{C} = -3.78^\circ\text{C}$

Practice Problems

• Problem 1: How many moles of ethylene glycol (C2H6O2) must be added to 1,000 g of water to form a solution that has a freezing point of -10°C?

• Problem 2: An ethylene glycol solution contains 28.3 g of ethylene glycol in 97.2 mL of water. Calculate the freezing point of the solution. (Density of water = 1.00 g/mL)

• Problem 3: When 825 g of an unknown is dissolved in 3.45 L of water, the freezing point of the solution is decreased by 2.89°C. Assuming the unknown compound is a non-electrolyte, calculate its molar mass.

Summary Table: Freezing Point Depression Variables

Variable	Definition
$\Delta T_f$	Freezing point depression (°C)
$K_f$	Cryoscopic constant (°C·kg/mol)
$m$	Molality (mol solute/kg solvent)
$i$	Van't Hoff factor (number of particles per formula unit)

Applications

• Antifreeze in car radiators uses ethylene glycol to lower the freezing point of water, preventing engine damage in cold climates.

• Salting roads in winter lowers the freezing point of water, helping to melt ice.