Colligative Properties: Raoult’s Law, Boiling/Freezing Point Elevation/Depression, Molality, Osmotic Pressure

Topic: mixtures and colligative properties (depend on solute concentration, not solute identity)
Four main colligative properties: vapor pressure lowering, boiling point elevation, freezing point depression, osmotic pressure (π)
Raoult's Law (vapor pressure lowering): the solvent’s vapor pressure in solution is reduced in proportion to the solvent’s mole fraction
Key definitions and quantities:
- P^0_solvent: vapor pressure of the pure solvent at a given temperature
- P_solution: vapor pressure of the solvent in the solution
- X_solvent: mole fraction of the solvent in the solution
- nsolvent, nsolute: moles of solvent and solute
- i: Van't Hoff factor (accounts for particle dissociation/association; i = 1 for nonelectrolytes; varies for electrolytes depending on dissociation)
Raoult’s Law for an ideal solution: $P{solution} = X{solvent} \, P^0_{solvent}$
- Xsolvent = $\frac{n{solvent}}{n{solvent} + n{solute}}$
- For electrolytes, the effective number of solute particles increases by i, so the mole fraction of solvent becomes $X{solvent} = \frac{n{solvent}}{n{solvent} + i \, n{solute}}$
How to obtain P^0_solvent:
- P^0_solvent comes from tabulated values for the substance at the given temperature; you look up the vapor pressure of the pure solvent
- The calculation assumes ideal behavior and Raoult’s law applies
Important notes about i (Van’t Hoff factor):
- Nonelectrolyte: i ≈ 1
- Strong electrolytes (fully dissociate in water): i ≈ number of particles produced; example: NaCl → Na^+ + Cl^- gives i ≈ 2
- HCl (strong acid): i ≈ 2 (H^+ + Cl^-)
- HF (weak acid): i ≈ 1 (partial dissociation) for practical purposes in many problems
- If the solvent is nonaqueous, i ≈ 1 (Raoult’s law for the solvent assumes ideal behavior)
Example 1: nonvolatile nonelectrolyte (glucose) dissolved in water
- Given: 50 g glucose (C6H12O6, M ≈ 180.16 g/mol) and 50 g water
- Moles: $n{solute} = \frac{50}{180.16} \approx 0.278\,\text{mol}$ , $n{solvent} = \frac{50}{18.015} \approx 2.775\,\text{mol}$
- X_solvent = $\frac{2.775}{2.775 + 0.278} \approx 0.909$
- P^0solvent at 25°C for water: $P^0{H2O} = 23.8\ \text{torr}$
- P_solution = $0.909 \times 23.8 \approx 21.6\ \text{torr}$
- Mole fraction is always between 0 and 1; if outside, calculation is likely wrong
Example 2: nonvolatile electrolyte (strong electrolyte) in water
- Example solute: a strong electrolyte that dissociates into multiple ions (e.g., HCl → H^+ + Cl^-; i ≈ 2)
- For a given solute amount, use effective moles of solute: i × n_solute
- Xsolvent = $\frac{n{solvent}}{n{solvent} + i\, n{solute}}$
- Then Psolution = $X{solvent} P^0_{solvent}$
Effect of temperature and intermolecular forces on vapor pressure
- Increasing temperature increases vapor pressure of the pure solvent
- Stronger intermolecular forces in solvent/solution lower vapor pressure of the solvent in solution
Phase diagram intuition: comparing a pure substance vs. a solution
- Freezing point depression: solute lowers the freezing point of the solvent
- Boiling point elevation: solute raises the boiling point of the solvent
- The dotted lines on a phase diagram illustrate the shifts in freezing/boiling points due to solute addition
Boiling point elevation (ΔT_b)
- Definition: the rise in boiling temperature of a solvent when a nonvolatile solute is added
- Formula: $\Delta Tb = i\, m\, Kb$
- m is molality: $m = \frac{n{solute}}{m{solvent}} = \frac{n{solute}}{m{solvent\, (kg)}}$
- Why molality, not molarity? Molarity varies with temperature (volume changes), but molality uses kilograms of solvent and is temperature-independent
- Units: molality units are $\text{mol} \cdot \text{kg}^{-1}$
Example 3: boiling point of a Zn(NO3)2 solution
- Given: 25.0 g Zn(NO3)2 in 150.0 g water
- Zn(NO3)2 dissociates into 3 ions: Zn^{2+} and 2 NO3^−, so i = 3
- Molar mass: Zn(NO3)2 ≈ 189.4 g/mol
- Moles solute: $n_{solute} = \frac{25.0}{189.4} \approx 0.132\,\text{mol}$
- Mass solvent: $m_{solvent} = 0.150\,\text{kg}$
- Molality: $m = \frac{0.132}{0.150} \approx 0.88\,\text{m}$
- Boiling point of pure solvent (water): $T_{BP}^{pure} = 100.0^{\circ}C$
- With water as solvent and K_b(H2O) ≈ 0.512\,°C·kg·mol^{-1}
- Change in boiling point: $\Delta Tb = i m Kb = 3 \times 0.88 \times 0.512 \approx 1.35^{\circ}C$
- Boiling point of the solution: $T_{BP}^{solution} \approx 100.0^{\circ}C + 1.35^{\circ}C \approx 101.35^{\circ}C$
Freezing point depression (ΔT_f)
- Definition: the freezing point of a solvent decreases upon addition of a nonvolatile solute
- Formula: $\Delta Tf = i\, m\, Kf$
- Kf (cryoscopic constant) for water: Kf \approx 1.86\,\frac{^{\circ}C}{\text{kg·mol}^{-1}}
- Sign convention: freezing point lowers by ΔT_f; often reported as a negative shift
Osmotic pressure (π)
- Definition: pressure required to prevent solvent flow across a semipermeable membrane due to solute presence
- Ideal relation (for dilute solutions): $\pi = i\, M \, R \, T$ where M is molarity, R is the gas constant, and T is temperature
- Note: π increases with solute concentration and temperature; i accounts for dissociation
Key equations (summary):
- Vapor pressure lowering: $P{solution} = X{solvent} \; P^0{solvent}$ with $X{solvent} = \frac{n{solvent}}{n{solvent} + i\, n_{solute}}$
- Boiling point elevation: $\Delta Tb = i\, m\, Kb$
- Freezing point depression: $\Delta Tf = i\, m\, Kf$
- Molality: $m = \frac{n{solute}}{m{solvent}}$ (mol/kg)
- Molarity vs Molality: M = mol solute / L of solution; m uses mass of solvent to stay temperature-independent
- Osmotic pressure: $\pi = i\, M \, R \, T$
Practical considerations and pitfalls
- Always check if solute dissociates; determine i accordingly
- For electrolytes with incomplete dissociation (weak electrolytes), i is less than the integer value, often requiring experimental data or dissociation fraction
- If solvent is nonaqueous, often treat i ≈ 1 for Raoult’s law simplification
- When calculating Xsolvent, remember that nsolvent and nsolute are in moles; Xsolvent is dimensionless and must lie between 0 and 1
Real-world relevance and connections
- Applications: antifreeze (ethylene glycol in water increases boiling point, lowers freezing point), food preservation, kidney dialysis, formulation of medicines, de-icing, and industrial separation processes
- Foundations: builds on colligative-property concepts from thermodynamics and solution chemistry; connects to Raoult’s law, ideal solutions, and dissociation chemistry
- Ethical/practical implications: correct handling of electrolytes in solutions affects safety (e.g., concentrated acids/bases), environmental impact of antifreeze compounds, and accurate formulation in pharmaceuticals
Quick recap of the workflow to solve typical problems
- Identify whether solute is volatile/nonvolatile and electrolyte/non-electrolyte
- Compute nsolute and nsolvent (use molar masses) and determine i
- For vapor pressure: compute Xsolvent with i, then Psolution with Raoult’s law
- For boiling/freezing points: compute m (molality) using mass of solvent; apply ΔTb or ΔTf with i, Kb or Kf
- For osmotic pressure: compute π using M, i, T, and R
Common exam-style tips
- Always report the sign of ΔT_f (negative for freezing point depression)
- Use kilograms for solvent mass when calculating molality
- Use tabulated Kb and Kf values for the solvent in question
- When solute dissociates, multiply nsolute by i to get the effective particle count for Xsolvent and for m in ΔTb/ΔTf
Provided numerical example steps (condensed):
- For 25.0 g Zn(NO3)2 in 150.0 g water: Molar mass Zn(NO3)2 ≈ 189.4 g/mol
- nsolute ≈ 25.0 / 189.4 ≈ 0.132 mol; msolvent ≈ 0.150 kg
- i = 3; m ≈ 0.132 / 0.150 ≈ 0.88 m
- ΔTb ≈ i m Kb = 3 × 0.88 × 0.512 ≈ 1.35 °C
- T_BP(solution) ≈ 100.0 °C + 1.35 °C ≈ 101.35 °C