Elevation of boiling point and depression of freezing point
A non-volatile solute lowers the vapour pressure, so a solution boils higher and freezes lower than its solvent: ΔTb = Kb m and ΔTf = Kf m.
Why the boiling point rises
A liquid boils when its vapour pressure equals the atmospheric pressure (1.013 bar). A non-volatile solute lowers the vapour pressure, so the solution must be heated to a higher temperature before its vapour pressure reaches 1.013 bar. For a dilute solution, the rise ΔTb is proportional to the molality: ΔTb = Kb m. Kb is the molal elevation constant (ebullioscopic constant), in K kg mol⁻¹.
Why the freezing point falls
At the freezing point, the liquid and the solid have the same vapour pressure. A solution has a lower vapour pressure than the pure liquid, so its vapour pressure curve meets the solid's curve at a lower temperature. For a dilute solution, ΔTf = Kf m, where Kf is the molal depression constant (cryoscopic constant). Kb and Kf depend only on the solvent: on its molar mass, its boiling or freezing point and its enthalpy of vaporisation or fusion.
How to draw the vapour pressure graphs
- Boiling point: draw temperature on the x-axis and vapour pressure on the y-axis. Draw a rising curve for the pure solvent and a second rising curve just below it for the solution.
- Draw a horizontal line at 1.013 bar. Mark where it cuts the solvent curve as Tb° and where it cuts the solution curve as Tb, to the right. Label the gap ΔTb.
- Freezing point: on similar axes, draw a steep curve for the solid solvent, the curve for the liquid solvent, and the solution curve below the liquid solvent curve.
- Mark where the solid curve meets the liquid solvent curve as Tf° and where it meets the solution curve as Tf, to the left. Label the gap ΔTf.
- Label every curve: solvent, solution, solid solvent.
Kb and Kf of some solvents
| Solvent | Kb (K kg mol⁻¹) | Kf (K kg mol⁻¹) |
|---|---|---|
| Water | 0.52 | 1.86 |
| Ethanol | 1.20 | 1.99 |
| Benzene | 2.53 | 5.12 |
| Cyclohexane | 2.79 | 20.00 |
Worked example: antifreeze in a radiator
Ethylene glycol (HOCH₂CH₂OH, M = 62 g mol⁻¹) is added to car radiator water as antifreeze. Find the freezing point of a solution of 62 g of ethylene glycol in 500 g of water. Kf of water = 1.86 K kg mol⁻¹; freezing point of water = 273.15 K.
- n₂ = 62 g ÷ 62 g mol⁻¹ = 1.0 mol; w₁ = 500 g = 0.500 kg.
- m = 1.0 mol ÷ 0.500 kg = 2.0 mol kg⁻¹.
- ΔTf = Kf × m = 1.86 K kg mol⁻¹ × 2.0 mol kg⁻¹ = 3.72 K.
- Tf = 273.15 K − 3.72 K = 269.43 K.
Tf = 269.43 K (about −3.7 °C)
Worked example: molar mass from boiling and freezing data
(a) 6.0 g of a non-volatile non-electrolyte in 200 g of water raises the boiling point by 0.26 K. Find its molar mass (Kb = 0.52 K kg mol⁻¹). (b) 0.64 g of naphthalene in 25 g of benzene lowers the freezing point by 1.024 K. Find the molar mass of naphthalene (Kf of benzene = 5.12 K kg mol⁻¹).
- Formula: M₂ = (K × w₂ × 1000) ÷ (ΔT × w₁).
- (a) M₂ = (0.52 × 6.0 × 1000) ÷ (0.26 × 200) = 3120 ÷ 52 = 60 g mol⁻¹.
- (b) M₂ = (5.12 × 0.64 × 1000) ÷ (1.024 × 25) = 3276.8 ÷ 25.6 = 128 g mol⁻¹.
- Check (b): naphthalene is C₁₀H₈, and 10 × 12 + 8 × 1 = 128 g mol⁻¹.
(a) 60 g mol⁻¹ (urea has this molar mass); (b) 128 g mol⁻¹
On winter drives to Leh or Tawang, ethylene glycol antifreeze in the radiator keeps the coolant liquid below 0 °C. Salt added to cooking water raises its boiling point only a little, a fraction of a degree.
2 marks: “Why is the freezing point of a solution lower than that of the pure solvent?” A liquid freezes when its vapour pressure equals that of its solid. The solute lowers the vapour pressure of the liquid, so this equality is reached at a lower temperature.
3 marks: “Find the boiling point of a solution of 3.42 g of sucrose (M = 342 g mol⁻¹) in 100 g of water. Kb = 0.52 K kg mol⁻¹; boiling point of water = 373.15 K.” n = 3.42 ÷ 342 = 0.01 mol; m = 0.01 ÷ 0.100 = 0.1 mol kg⁻¹. ΔTb = 0.52 × 0.1 = 0.052 K. Tb = 373.15 + 0.052 = 373.202 K, about 373.20 K.
5 marks: “Define elevation of boiling point and molal elevation constant. Draw a graph to show the elevation of boiling point, and derive the formula for the molar mass of a solute from it.” ΔTb = Tb − Tb°; Kb is the elevation for a 1 molal solution. Graph: vapour pressure against temperature, solution curve below the solvent curve, each cutting the 1.013 bar line; ΔTb between them. ΔTb = Kb × (w₂ × 1000) ÷ (M₂ × w₁), so M₂ = (Kb × w₂ × 1000) ÷ (ΔTb × w₁).
Water's two numbers: 0.52 for boiling, 1.86 for freezing. A solute pushes the boiling point up and the freezing point down.