Freezing Point Depression — Worked Problems and Formula
Salt spread on an icy road, antifreeze in a car radiator, and the classic laboratory method for finding the molar mass of an unknown solid are all the same piece of chemistry. Adding a non-volatile solute lowers the freezing point of a solvent, and the size of that drop depends only on how many solute particles are present — not on what they are. This guide works through four problems of the types that actually appear in board and entrance papers.
The formula
| Symbol | Meaning | Unit |
|---|---|---|
| ΔTf | depression in freezing point (always a positive number) | K or °C (the size of a degree is the same in both) |
| Kf | cryoscopic constant — a property of the solvent only | K kg mol⁻¹ |
| m | molality = moles of solute ÷ kilograms of solvent | mol kg⁻¹ |
| i | van 't Hoff factor — particles produced per formula unit | no unit |
For finding a molar mass, substitute m = (w₂ ÷ M₂) ÷ (w₁ ÷ 1000) and rearrange:
Here w₂ is the mass of solute in grams, w₁ the mass of solvent in grams, and the 1000 converts grams of solvent to kilograms. Every symbol in that formula is a number you either measure or look up.
Cryoscopic constants worth knowing
| Solvent | Freezing point (°C) | Kf (K kg mol⁻¹) |
|---|---|---|
| Water | 0.0 | 1.86 |
| Benzene | 5.5 | 5.12 |
| Camphor | 179 | 39.7 |
| Cyclohexane | 6.5 | 20.0 |
Camphor's enormous Kf is why Rast's method uses it: a tiny amount of solute produces a depression of several degrees, which even a simple thermometer can read accurately. Different data tables round these values slightly differently — use the value printed in your question.
Worked problem 1 — a straightforward depression
Problem: 5.00 g of glucose (C₆H₁₂O₆) is dissolved in 100 g of water. Find the freezing point of the solution. Kf(water) = 1.86 K kg mol⁻¹.
Step 1 — molar mass of glucose.
6 × 12.011 + 12 × 1.008 + 6 × 15.999 = 72.066 + 12.096 + 95.994 = 180.16 g/mol
Step 2 — moles of solute. n = 5.00 ÷ 180.16 = 0.02775 mol
Step 3 — molality. Solvent = 100 g = 0.100 kg
m = 0.02775 ÷ 0.100 = 0.2775 mol/kg
Step 4 — depression. Glucose is a non-electrolyte, so i = 1.
ΔTf = 1 × 1.86 × 0.2775 = 0.516 K
Step 5 — the freezing point itself.
Tf = 0.0 − 0.516 = −0.516 °C
Step 5 is where marks are lost. ΔTf is the drop; the question asked for the freezing point, so you must subtract.
Worked problem 2 — molar mass by the cryoscopic method
Problem: 1.50 g of a non-volatile, non-electrolyte solute is dissolved in 30.0 g of benzene. The freezing point of the benzene falls by 2.56 K. Kf(benzene) = 5.12 K kg mol⁻¹. Find the molar mass of the solute.
Using M₂ = (1000 · Kf · w₂) ÷ (ΔTf · w₁), with i = 1:
Numerator: 1000 × 5.12 × 1.50 = 5120 × 1.50 = 7680
Denominator: 2.56 × 30.0 = 76.8
M₂ = 7680 ÷ 76.8 = 100 g/mol
Check it backwards. If M₂ = 100, then n = 1.50 ÷ 100 = 0.0150 mol in 0.0300 kg of benzene, so m = 0.0150 ÷ 0.0300 = 0.500 mol/kg, and ΔTf = 5.12 × 0.500 = 2.56 K. ✓ It reproduces the given data exactly.
Always do a reverse check like this on a molar-mass answer. It takes fifteen seconds and catches a misplaced factor of 1000 immediately.
Worked problem 3 — an electrolyte, and the degree of dissociation
Problem: 1.00 g of NaCl is dissolved in 100 g of water. (a) Predict the freezing point assuming complete dissociation. (b) The measured freezing point is −0.614 °C. Find the van 't Hoff factor and the degree of dissociation.
(a) Step 1 — moles. M(NaCl) = 22.990 + 35.45 = 58.44 g/mol
n = 1.00 ÷ 58.44 = 0.017112 mol
Step 2 — molality. m = 0.017112 ÷ 0.100 = 0.17112 mol/kg
Step 3 — with i = 2.
ΔTf = 2 × 1.86 × 0.17112 = 2 × 0.31828 = 0.6366 K
Predicted freezing point = −0.637 °C
(b) From the measurement. Observed ΔTf = 0.614 K.
i = observed ΔTf ÷ ΔTf for i = 1
i = 0.614 ÷ 0.31828 = 1.93
Degree of dissociation. For a salt giving n particles,
i = 1 + α(n − 1). With n = 2:
α = i − 1 = 1.93 − 1 = 0.93, i.e. 93% dissociated
Ion pairing — a Na⁺ and a Cl⁻ moving together for part of the time — is why the real figure falls short of 100%. The effect grows as concentration rises, which is exactly why colligative formulas are stated for dilute solutions.
Worked problem 4 — why roads are salted, and with what
Compare three solutes at the same 1.00 mol/kg concentration in water, assuming ideal dissociation:
| Solute | Particles per formula unit (i) | ΔTf = i × 1.86 × 1.00 | Freezing point |
|---|---|---|---|
| Glucose C₆H₁₂O₆ | 1 | 1.86 K | −1.86 °C |
| Sodium chloride NaCl | 2 | 3.72 K | −3.72 °C |
| Calcium chloride CaCl₂ | 3 | 5.58 K | −5.58 °C |
Per mole, CaCl₂ is three times as effective as glucose because it releases three particles. That, together with the fact that dissolving CaCl₂ is exothermic and so releases a little heat of its own, is why calcium chloride is preferred where temperatures fall well below freezing. Ethylene glycol in a car radiator works on the same principle, with the added advantage that it is a liquid that mixes with water in any proportion.
Common mistakes that cost marks
- Using molarity instead of molality. ΔTf needs moles per kilogram of solvent, not per litre of solution. They are close for dilute aqueous solutions and quite different otherwise.
- Dividing by grams of solvent instead of kilograms. An answer 1000 times too small — the single most frequent slip in this topic.
- Using the mass of the solution as the mass of solvent. w₁ is the solvent alone.
- Reporting ΔTf when the question asked for the freezing point, or forgetting that the solvent may not freeze at 0 °C (benzene freezes at 5.5 °C, so a depression of 2.56 K gives 2.94 °C, not −2.56 °C).
- Leaving out i for an ionic solute, or applying i to glucose, urea and sucrose, which do not dissociate.
- Treating Kf as a property of the solute. It belongs to the solvent. Change the solvent and Kf changes; change the solute and it does not.
- Assuming the solute dissolves in the solid solvent too. The whole effect depends on the solute staying out of the ice crystal that forms.
Where this appears in exams
| Exam | Typical question |
|---|---|
| CBSE/ICSE Class 12 | Molar mass by cryoscopy; freezing point of a salt solution; van 't Hoff factor from data |
| Class 12 practical | Determining the molar mass of an unknown solid using a Beckmann-type experiment |
| JEE/NEET | Ranking solutions by freezing point; abnormal molar masses from association or dissociation |
| IIT-JAM / CUET-PG | Deriving Kf from the enthalpy of fusion; degree of dissociation problems |
| GATE / CSIR-NET | Activity of the solvent, non-ideal solution corrections |
The molality step is where most freezing-point answers go wrong. Get the concentration right and the rest is one multiplication. The Concentration tool converts between mass, moles and concentration for any solute, so you can confirm the number you feed into ΔTf = i Kf m before you commit to it.
Open the Concentration / Molarity Calculator →If colligative-property numericals are costing you marks in tests, the fix is usually structured practice rather than more theory. ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre and online classes across India, and for students in Delhi, Noida and Gurgaon who want one-to-one help at home, home tuition is available through delhihometutor.com. Course details: abcchemistry.in.