Colligative Properties

Colligative properties are dependent on the concentration of solute particles, not the identity of those particles. The main colligative properties discussed include:

• Boiling Point Elevation
• Freezing Point Depression
• Osmotic Pressure
• Vapor Pressure

1. Boiling Point Elevation

• Definition: Adding solute particles raises the boiling point of a solution.
• Example: Pure water boils at 100°C at 1 atm. Adding salt increases this boiling point.
  • 1 M NaCl solution: ~101°C
  • 5 M NaCl solution: ~105.1°C
  • 10 M NaCl solution: ~110°C
• Formula: ΔT = Kb x m x i
  • Kb = boiling point elevation constant
  • m = molality
  • i = van't Hoff factor
• Calculation Example: For a 5m solution, ΔT = 0.51 x 5 x 2 = 5.1, so boiling point = 105.1°C.

2. Freezing Point Depression

• Definition: Adding solute particles lowers the freezing point of a solution.
• Practical Use: Salt is used on roads in winter to prevent ice formation by lowering the freezing point.
• Formula: ΔT = -Kf x m x i
  • Kf = freezing point depression constant
• Example: 1m NaCl solution lowers freezing point by 3.72°C.
  • For a 5m AlCl3 solution, change = -37.2°C.

3. Vapor Pressure

• Dependence: Vapor pressure decreases with increased solute concentration.
• Formula: P_solution = X_solvent x P°_solvent
  • X_solvent = mole fraction of the solvent
• Explanation: Increasing solute lowers mole fraction of solvent, reducing vapor pressure._

4. Osmotic Pressure

• Formula: π = M x R x T x i
  • M = molarity
  • R = gas constant
  • T = temperature in Kelvin
• Dependence: Osmotic pressure increases with higher solute concentration.

Summary of Relationships

• Direct Relationships:
  • Boiling Point ↑ with solute concentration ↑
  • Osmotic Pressure ↑ with molarity ↑
• Inverse Relationships:
  • Freezing Point ↓ with solute concentration ↑
  • Vapor Pressure ↓ with solute concentration ↑

Example Problems

1. Boiling Point Calculation:

   • 20g of NaOH in 200g of water:
     • Molality = moles of solute / kg of solvent
     • Boiling point = 102.55°C

2. Freezing Point Calculation:

   • 400g AlCl3 in 1600g of water:
     • Molality = moles of solute / kg of solvent
     • Freezing point = -13.95°C

3. Comparing Boiling Points:

   • Highest boiling point determined by highest m x i value.
   • Example: 0.8m NaCl (i=2) → highest boiling point in given options.