§ C-31 Raoult’s Law & Colligative Properties: Ebullioscopy ΔT_b & Cryoscopy ΔT_f (Grade 9–11 · Colligative Properties)
Chapter XXXI. Vapor Pressure Lowering ΔP = χ₂P°, Boiling Point Elevation & Freezing Point Depression. «The lowering of vapor pressure, elevation of boiling point, and depression of freezing point depend solely on the concentration of solute particles.», François-Marie Raoult, 1887
Theory Summary & Scientific Mechanism
Colligative properties depend solely on the total concentration of dissolved solute particles (i· m, where i is the van ’t Hoff dissociation factor and m is molality) rather than their chemical identity.
By Raoult’s Law, a non-volatile solute reduces the surface mole fraction of solvent (χ₁ = 1 - χ₂), lowering the equilibrium vapor pressure proportionally: P = χ₁ P₀^° ⇒ Δ P = P₀^° - P = χ₂ P₀^°.
Because the solution’s vapor pressure curve is depressed relative to pure solvent, the boiling point rises (ebullioscopy, Δ T_b = i K_b m, K_b = 0.512 K·kg/mol for water) and the freezing point drops (cryoscopy, Δ T_f = i K_f m, K_f = 1.86 K·kg/mol), explaining road de-icing salts and antifreeze coolants.
Everyday Analogy
Dissolved salt ions act like friendly blockers at the liquid surface, making it harder for water molecules to escape into vapor or lock into ice. That’s why salted water boils above 100 ^°C and melts road ice below 0 ^°C!
Interactive Question & Derivation
Why does spreading salt (NaCl, i ≈ 2 or CaCl₂, i ≈ 3) melt ice on winter roads, and why does an aqueous solution boil at a temperature above 100 ^°C?
- A: Non-volatile solute particles lower the solvent’s saturated vapor pressure by Raoult’s law (Δ P = χₛolute P₀^°), shifting freezing down by Δ T_f = i K_f m and boiling up by Δ T_b = i K_b m
- B: Dissolving salt chemically reacts with ice to generate combustion flames that raise water’s vapor pressure
- C: Colligative properties depend only on the color of the solute and are independent of molality m or the van ’t Hoff factor i
Study Hint: By Raoult’s law, a non-volatile solute lowers vapor pressure (Δ P = χ₂ P₀^°), raising T_b by i K_b m and depressing T_f by i K_f m.
Proof & Explanation: According to Raoult’s Law, a non-volatile solute lowers the mole fraction of solvent at the surface and reduces its chemical potential, depressing the equilibrium vapor pressure: P = χ₁ P₀^° ⇒ Δ P = P₀^° - P = χ₂ P₀^°. Because the solution’s vapor pressure curve lies below that of pure water, a higher temperature is needed to reach 1 atm (boiling point elevation Δ T_b = i K_b m, K_b = 0.512 K·kg/mol) and a lower temperature is required to freeze (freezing point depression Δ T_f = i K_f m, K_f = 1.86 K·kg/mol).
Key Equations & 3D Simulation Parameters
P = χ₁ P₀^°, Δ T_b = i K_b m, Δ T_f = i K_f m, χ₂ = (i m)/(55.5 + i m)
- Solute molality m (moles per 1 kg H₂O): 0.5 … 5 mol/kg (default: 2 mol/kg)
- Van ’t Hoff factor i (1=glucose, 2=NaCl, 3=CaCl₂): 1 … 3 (default: 2)
- Pure solvent vapor pressure P₀°: 3.2 … 101.3 kPa (default: 23.8 kPa)
- Manometer bath temperature t: 25 … 100 °C (default: 65 °C)