Calculate freezing point depression, boiling point elevation, and osmotic pressure
Colligative Properties: Properties that depend only on the number of solute particles
• Freezing Point Depression: ΔTf = i·Kf·m
• Boiling Point Elevation: ΔTb = i·Kb·m
• Osmotic Pressure: π = i·M·R·T
mol solute / kg solvent
Number of particles per formula unit (1 for non-electrolytes, 2 for NaCl, etc.)
°C·kg/mol (1.86 for water)
| Solvent | Kf (°C·kg/mol) | Kb (°C·kg/mol) |
|---|---|---|
| Water | 1.86 | 0.512 |
| Benzene | 5.12 | 2.53 |
| Camphor | 37.7 | 5.95 |
| Acetic Acid | 3.90 | 3.07 |
Colligative properties depend only on the number of solute particles, not their identity. They include:
Freezing Point Depression
ΔTf = i·Kf·m
Boiling Point Elevation
ΔTb = i·Kb·m
Osmotic Pressure
π = i·M·R·T
Number of particles produced per formula unit:
• Non-electrolytes (sugar, urea): i = 1
• NaCl, MgSO₄: i = 2
• CaCl₂, Na₂SO₄: i = 3
• AlCl₃: i = 4
Problem:
What is the freezing point of a solution made by dissolving 58.5 g of NaCl (1.0 mol) in 1.0 kg of water?
Solution:
1. Molality: m = 1.0 mol / 1.0 kg = 1.0 m
2. van't Hoff factor for NaCl: i = 2 (Na⁺ + Cl⁻)
3. Kf for water = 1.86 °C·kg/mol
4. ΔTf = i·Kf·m = (2)(1.86)(1.0)
5. ΔTf = 3.72°C
6. New freezing point = 0°C - 3.72°C
Freezing Point = -3.72°C
This is why salt is used to melt ice on roads!
Problem:
Calculate the boiling point of a solution containing 0.5 mol of sugar (C₁₂H₂₂O₁₁) in 1.0 kg of water.
Solution:
1. Molality: m = 0.5 mol / 1.0 kg = 0.5 m
2. van't Hoff factor for sugar: i = 1 (non-electrolyte)
3. Kb for water = 0.512 °C·kg/mol
4. ΔTb = i·Kb·m = (1)(0.512)(0.5)
5. ΔTb = 0.256°C
6. New boiling point = 100°C + 0.256°C
Boiling Point = 100.256°C
The effect is small but measurable in cooking
Isotonic Solution (0.9% NaCl): Cell maintains shape
Hypotonic (pure water): Cell swells and bursts (hemolysis)
Hypertonic (high salt): Cell shrinks (crenation)
Turgor pressure from osmosis keeps plants rigid. When water is scarce, cells lose water (plasmolysis) and the plant wilts.
Osmotic pressure of 0.15 M glucose at 37°C (body temperature):
π = i·M·R·T
π = (1)(0.15 M)(0.0821 L·atm/(mol·K))(310 K)
π = 3.82 atm
This significant pressure drives water movement across cell membranes!