BackFreezing Point Depression in Solutions
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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:
= change in freezing point (°C)
= van't Hoff factor (number of particles per formula unit)
= freezing point depression constant (°C·kg/mol)
= 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:
Mass of water: 302.6 g = 0.3026 kg
Molality:
For glucose, (non-electrolyte)
Freezing point of solution:
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 |
|---|---|
Freezing point depression (°C) | |
Cryoscopic constant (°C·kg/mol) | |
Molality (mol solute/kg solvent) | |
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.