Colligative Properties
Properties that depend only on the number of solute particles, not their identity
The Four Colligative Properties
1. Vapor Pressure Lowering
Psolution = χsolvent × P°solvent
Raoult's Law
ΔP = P° - P = χsolute × P°
χ = mole fraction, P° = vapor pressure of pure solvent
2. Boiling Point Elevation
ΔTb = Kb × m × i
Where:
- • ΔTb = boiling point elevation (°C)
- • Kb = ebullioscopic constant (°C·kg/mol)
- • m = molality (mol/kg)
- • i = van't Hoff factor (# of particles per formula unit)
Kb(H₂O) = 0.512 °C·kg/mol
3. Freezing Point Depression
ΔTf = Kf × m × i
Where:
- • ΔTf = freezing point depression (°C)
- • Kf = cryoscopic constant (°C·kg/mol)
- • m = molality (mol/kg)
- • i = van't Hoff factor
Kf(H₂O) = 1.86 °C·kg/mol
4. Osmotic Pressure
Π = MRT × i
Where:
- • Π = osmotic pressure (atm)
- • M = molarity (mol/L)
- • R = 0.0821 L·atm/(mol·K)
- • T = temperature (K)
- • i = van't Hoff factor
Van't Hoff Factor (i)
| Solute Type | Example | i (ideal) | Explanation |
|---|---|---|---|
| Non-electrolyte | C₆H₁₂O₆, C₁₂H₂₂O₁₁ | 1 | Doesn't dissociate |
| Strong electrolyte (1:1) | NaCl, HCl, KBr | 2 | 2 ions per formula unit |
| Strong electrolyte (1:2) | CaCl₂, Na₂SO₄ | 3 | 3 ions per formula unit |
| Strong electrolyte (2:3) | Al₂(SO₄)₃ | 5 | 2 Al³⁺ + 3 SO₄²⁻ |
| Weak electrolyte | CH₃COOH | 1 < i < 2 | Partial dissociation |
Worked Examples
Example 1: Freezing Point Depression
Problem: What is the freezing point of a solution containing 25.0 g of glucose (C₆H₁₂O₆) in 150 g of water?
Solution:
Step 1: Calculate moles of glucose
Molar mass = 180.16 g/mol
moles = 25.0 g ÷ 180.16 g/mol = 0.1388 mol
Step 2: Calculate molality
m = 0.1388 mol ÷ 0.150 kg = 0.925 mol/kg
Step 3: Calculate ΔTf
ΔTf = Kf × m × i
ΔTf = 1.86 °C·kg/mol × 0.925 mol/kg × 1
ΔTf = 1.72 °C
Freezing point = 0°C - 1.72°C = -1.72°C
Example 2: Boiling Point with Electrolyte
Problem: Calculate the boiling point of a solution of 10.0 g NaCl in 200 g of water.
Solution:
Step 1: Calculate molality
Molar mass NaCl = 58.44 g/mol
moles = 10.0 g ÷ 58.44 g/mol = 0.171 mol
m = 0.171 mol ÷ 0.200 kg = 0.855 mol/kg
Step 2: Determine i for NaCl
NaCl → Na⁺ + Cl⁻ (i = 2)
Step 3: Calculate ΔTb
ΔTb = 0.512 °C·kg/mol × 0.855 mol/kg × 2
ΔTb = 0.876 °C
Boiling point = 100°C + 0.876°C = 100.88°C
Example 3: Osmotic Pressure
Problem: Calculate the osmotic pressure of a 0.100 M glucose solution at 25°C.
Solution:
Π = MRT × i
Π = (0.100 mol/L)(0.0821 L·atm/mol·K)(298 K)(1)
Π = 2.45 atm
Π = 2.45 atm
Common Constants
Kb Values
Water: 0.512 °C·kg/mol
Benzene: 2.53 °C·kg/mol
Chloroform: 3.63 °C·kg/mol
Acetic acid: 3.07 °C·kg/mol
Kf Values
Water: 1.86 °C·kg/mol
Benzene: 5.12 °C·kg/mol
Cyclohexane: 20.0 °C·kg/mol
Camphor: 37.7 °C·kg/mol
Common Mistakes
Using Molarity Instead of Molality
ΔTb and ΔTf use molality (mol/kg), NOT molarity!
Forgetting van't Hoff Factor
Electrolytes dissociate! NaCl has i=2, not i=1.
Wrong Temperature Units
Osmotic pressure uses Kelvin, not Celsius!
Memory Aid
Colligative = collective effect of particles, regardless of identity