Freezing Point Depression and Boiling Point Elevation
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Direct answer
A dissolved non-volatile solute depresses the freezing point by ΔTf = Kf × m and raises the boiling point by ΔTb = Kb × m, with m the molality — kilograms of solvent, not litres of solution, because molality survives temperature change. Water's constants anchor the arithmetic: Kf = 1.86 and Kb = 0.52 kelvin kilogram per mol, so a 1-molal sugar solution freezes at −1.86 °C and boils at 100.52 °C. Electrolytes multiply the effect by the van't Hoff factor i — ideal 2 for sodium chloride, 3 for calcium chloride — so 1 molal sodium chloride ideally freezes near −3.72 °C. The same equations extract molar mass: M = (Kf × w2 × 1000)/(ΔTf × w1), with NCERT's own worked example landing at 256 g per mol for an unknown solute in benzene.
What you must remember
- Two equations: ΔTf = Kf × m and ΔTb = Kb × m, both proportional to molality; m = moles of solute per kg of solvent.
- Constants to carry: water Kf = 1.86, Kb = 0.52; benzene Kf = 5.12, Kb = 2.53 kelvin kg mol−1 — benzene's large Kf makes it the sensitive solvent for molar mass work.
- One-molal benchmarks: sugar water freezes at −1.86 °C, boils at 100.52 °C; ideal 1 m NaCl reaches about −3.72 °C through i = 2.
- Molar mass formula: M2 = (Kf × w2 × 1000)/(ΔTf × w1) — solute mass w2, solvent mass w1, both in grams; the Kb mirror image works for elevation data.
- van't Hoff factor: i = observed colligative effect ÷ calculated for no ionisation; strong electrolytes approach their ion count (NaCl 2, CaCl2 3, K2SO4 3), and the effective formula becomes ΔTf = i × Kf × m.
- Applications with reasons: ethylene glycol as engine antifreeze and salt on icy roads both exploit depression; adding salt to cooking water raises the boiling temperature a fraction of a degree through elevation.
- Why Kf exceeds Kb for water: Kf = RT(f)²M1/(1000 × ΔH(fus)) and Kb = RT(b)²M1/(1000 × ΔH(vap)); water's enormous enthalpy of vaporisation shrinks its Kb.
From a depression to a molar mass
NCERT's own numbers: 1.0 g of a non-electrolyte dissolved in 50 g of benzene lowers the freezing point by 0.40 K; Kf for benzene is 5.12. Molality = ΔTf/Kf = 0.40/5.12 = 0.078 mol per kg. That corresponds to 0.078 × 0.050 = 0.0039 mol in the 50 g taken, so the molar mass = 1.0/0.0039 ≈ 256 g per mol. Or in one substitution: M = (5.12 × 1.0 × 1000)/(0.40 × 50) = 256. Sulfur dissolved this way reads near 256 because S8 rings (8 × 32) dissolve as whole molecules — a structural deduction hanging off pure colligative arithmetic.
The electrolyte layer: the same 1 molal concentration as sodium chloride gives double the particles, hence double the depression ideally; observed values fall slightly short (i near 1.9 rather than 2) through interionic attraction, a refinement the theory section expects you to state. Ranking questions run on this: equimolal glucose, NaCl, CaCl2, AlCl3 depress freezing in rising order 1, 2, 3, 4 — a one-line comparison worth a mark most sessions.
Molality, i-factor and constant slips
Molality against molarity is the evergreen: if the question gives solution volume, convert through density to solvent mass before anything else. Forgetting i halves or thirds the expected depression — match the solute to its ion count before choosing a formula. Cross-pairing constants (Kb in a freezing problem) survives careless reading because the numbers look plausible. Two conditions must be verified before the plain formula applies: the solute must be non-volatile (or the "elevation" muddles into vapour pressure questions) and non-electrolytic (or multiply by i). And note what depression physically is — the solution's vapour pressure curve meets the solid's below the normal freezing point; the diagrams the theory section asks you to sketch in words.
Frequently asked questions
Why is molality used instead of molarity in colligative formulas?
Molality is mass-based and temperature-independent, while molarity changes as solution volume expands or contracts with temperature.
What are Kf and Kb for water?
1.86 and 0.52 kelvin kilogram per mol respectively — so 1 molal glucose freezes at −1.86 °C and boils at 100.52 °C.
Which freezes lower: 1 m glucose or 1 m sodium chloride?
Sodium chloride — ideally twice the depression (−3.72 °C) because each formula unit yields two particles.
A 1 g solute in 50 g benzene depresses freezing by 0.40 K (Kf 5.12) — molar mass?
M = (5.12 × 1 × 1000)/(0.40 × 50) = 256 g per mol, NCERT's benchmark result.
Why does spreading salt melt ice?
Dissolved salt depresses water's freezing point below the ambient temperature, so the ice-salt equilibrium shifts to liquid brine.