Chapter 13 · Solutions
13.9Solution Concentration and Freezing and Boiling
11 min · two checks
Predict
Why does salt water freeze below 0 °C and boil above 100 °C?
The idea
- Describe freezing-point depression and boiling-point elevation.
- Count particles from an ionic formula.
Colligative properties depend on the collection of solute particles, not on what the particles are (in the ideal case). Freezing-point depression: ΔT_f = i × K_f × m, where m is molality, moles of solute per kilogram of solvent, and i is the number of particles per formula unit. For NaCl, i is about 2. For sugar, i is 1. For CaCl₂, i is about 3. Boiling-point elevation has the same shape with K_b. K_f for water is 1.86 °C/m. K_b for water is 0.512 °C/m.
Molality is not molarity. Molality uses kilograms of solvent, not liters of solution, so it does not change with temperature the way a volume-based unit does. In dilute water solutions the numbers are similar and problems sometimes approximate. The direction to remember without a formula: solute lowers the freezing point and raises the boiling point. Road salt and engine antifreeze are the same idea.
Keep these
- ΔT_f = i K_f m. ΔT_b = i K_b m.
- i is the particle count: 1 for sugar, 2 for NaCl, 3 for CaCl₂.
- Molality is mol solute per kg solvent.
Worked path
Estimate the freezing point of a 0.50 m aqueous sugar solution. K_f = 1.86 °C/m. Sugar does not ionize.
Observe
i = 1. m = 0.50 mol/kg.