Colligative Properties in Antifreeze and De-icing — ΔTf = i·Kf·m in Practice
Freezing point depression is one of the first formulas where a syllabus equation maps directly onto something engineered at industrial scale. Engine coolant, aircraft de-icing fluid and the salt spread on winter roads are all the same chemistry: dissolve something in water and the water freezes lower. This article works that arithmetic properly for the three substances actually used, and then does the part most notes skip — showing by how much the ideal formula misses in the concentrated mixtures that real applications use.
The formula
with the molality defined as
What each term means
| Symbol | Meaning | Unit |
|---|---|---|
| ΔTf | Freezing point depression — a positive number you subtract from the pure solvent's freezing point | K (same size as °C) |
| Kf | Cryoscopic constant — a property of the solvent only. For water, 1.86 K·kg·mol−1 | K·kg·mol−1 |
| Kb | Ebullioscopic constant of the solvent. For water, 0.512 K·kg·mol−1 | K·kg·mol−1 |
| m | Molality — moles of solute per kg of solvent, not per litre of solution | mol·kg−1 |
| i | van't Hoff factor — the number of particles a formula unit actually produces in solution | dimensionless |
The word colligative means the effect depends only on how many solute particles there are, not on what they are. That is the whole reason a small, cheap, highly ionising salt is such an effective de-icer.
Worked example 1 — ethylene glycol coolant
Ethylene glycol is C2H6O2. Compute its molar mass first, from IUPAC atomic masses C = 12.011, H = 1.008, O = 15.999:
C: 2 × 12.011 = 24.022
H: 6 × 1.008 = 6.048
O: 2 × 15.999 = 31.998
M = 24.022 + 6.048 + 31.998 = 62.068 g/mol
Now take 500.0 g of glycol dissolved in 1.000 kg of water (roughly the 1 : 1 by mass mixture used as a starting point in practice).
moles = 500.0 ÷ 62.068 = 8.056 mol
molality m = 8.056 mol ÷ 1.000 kg = 8.056 mol/kg
glycol is a molecular solute that does not ionise, so i = 1
ΔTf = 1 × 1.86 × 8.056 = 14.98 K
Predicted freezing point = 0.00 − 14.98 = −15.0 °C
Hold on to that number. A real 50% by mass glycol–water mixture does not freeze at −15 °C; it stays liquid to a far lower temperature. The ideal formula has under-predicted the effect by a large margin, and the reason is explained in the limits section below. This is not a defect in your arithmetic — it is the formula being used outside the range it was derived for.
Worked example 2 — the boiling side of the same mixture
Coolant has two jobs: not freezing in winter and not boiling in summer. The same molality gives the elevation.
ΔTb = i × Kb × m = 1 × 0.512 × 8.056 = 4.12 K
Predicted boiling point at 1 atm = 100.00 + 4.12 = 104.1 °C
A cooling system is also pressurised, and raising the pressure raises the boiling point independently of any solute. The two effects add, which is why the chemistry alone does not explain the full operating temperature of a real system.
Worked example 3 — sodium chloride as a road de-icer
M(NaCl) = 22.990 + 35.45 = 58.44 g/mol
Take 100.0 g of NaCl in 1.000 kg of water.
moles = 100.0 ÷ 58.44 = 1.7112 mol
m = 1.7112 mol/kg
ideal i = 2 (one Na+ and one Cl−)
ΔTf = 2 × 1.86 × 1.7112 = 6.37 K
Predicted freezing point = −6.37 °C
Worked example 4 — calcium chloride, and the per-gram comparison
M(CaCl2) = 40.078 + 2 × 35.45 = 40.078 + 70.90 = 110.978 g/mol
Take the same 100.0 g in 1.000 kg of water.
moles = 100.0 ÷ 110.978 = 0.9011 mol
ideal i = 3 (one Ca2+ and two Cl−)
ΔTf = 3 × 1.86 × 0.9011 = 5.03 K
Predicted freezing point = −5.03 °C
This is a result worth pausing on. CaCl2 releases three particles per formula unit against NaCl's two, yet per gram it depresses the freezing point less — 5.03 K against 6.37 K. The heavier formula mass more than cancels the extra ion. If an exam question asks which is the better de-icer "per kilogram", the ideal colligative calculation says sodium chloride.
In practice calcium chloride is still chosen for the coldest conditions, for reasons that sit outside the colligative formula entirely: its eutectic temperature — the lowest freezing point any water–salt mixture of that pair can reach — is considerably lower than sodium chloride's, and its dissolution releases heat rather than absorbing it. Neither of those facts is in ΔTf = i·Kf·m. The formula answers one question well and is silent on the others.
| Solute (100.0 g per 1.000 kg water) | M (g/mol) | moles | ideal i | ΔTf (K) |
|---|---|---|---|---|
| Ethylene glycol C2H6O2 | 62.068 | 1.6111 | 1 | 3.00 |
| Sodium chloride NaCl | 58.44 | 1.7112 | 2 | 6.37 |
| Calcium chloride CaCl2 | 110.978 | 0.9011 | 3 | 5.03 |
(The glycol row recomputed at 100.0 g for a fair comparison: 100.0 ÷ 62.068 = 1.6111 mol, then 1 × 1.86 × 1.6111 = 3.00 K.) Ionising solutes win comfortably per gram — that is the colligative principle doing exactly what it should.
Where the simple formula stops being valid
- ΔTf = i·Kf·m is a dilute-solution limiting law. It is derived assuming an ideal solution. A 50% by mass coolant is nowhere near dilute, so example 1's −15 °C is not the real freezing point of that mixture — real glycol–water mixtures stay liquid far below it. Use the formula to understand the trend and to answer exam questions; never use it to specify a real coolant.
- i is almost never the whole number you wrote down. At the concentrations used on roads, ion pairing means the effective van't Hoff factor for NaCl is measurably below 2, so the actual depression is smaller than the ideal calculation. The ideal i is an upper bound, not a value.
- There is a floor: the eutectic. Adding more salt does not depress the freezing point indefinitely. Past the eutectic composition, extra salt simply will not dissolve and the mixture freezes at the eutectic temperature. The linear formula has no such limit built in, so extrapolating it predicts freezing points that cannot occur.
- Molality, not molarity. The denominator is the mass of solvent, not the volume or mass of the whole solution. Using 1.000 kg of solution in example 1 instead of 1.000 kg of water changes the answer substantially.
- Kf belongs to the solvent. 1.86 is water's value. It has nothing to do with the solute, and a question that gives you a different solvent gives you a different Kf.
- Depression means subtract. ΔTf is quoted positive; the freezing point is 0 − ΔTf. Reporting "+15 °C" is a real and frequent slip.
- De-icing is a surface-melting process, not an equilibrium calculation. Salt on a road has to dissolve into a thin brine layer first, and how fast that happens at low temperature is a kinetics question the colligative formula does not address.
Where this appears in exams
| Exam | Typical question |
|---|---|
| IIT-JAM | Compute ΔTf or molar mass from a measured depression; compare two electrolytes at equal mass or equal molality |
| CUET-PG | Direct substitution, and identifying i for a given electrolyte |
| GATE | Non-ideality, activity coefficients, and why the observed i differs from the formula value |
| CSIR-NET | Thermodynamic origin of colligative properties from chemical potential and solvent activity |
Check the arithmetic. These problems are lost on molar masses and unit conversions far more often than on the physics. The calculator suite includes the Molar Mass & Composition tool for the first step and a full scientific calculator for the rest — there is no dedicated freezing-point-depression tool, so the honest link is to the suite itself.
Open the ABC Chemistry Calculator Suite →Preparing for IIT-JAM, GATE, CSIR-NET or CUET-PG? ABC Chemistry runs dedicated competitive-exam batches at its coaching centre and online for students across India — details at abcchemistry.in.