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Colligative property problems look like one formula, ΔT = iKm, applied over and over. Marks are lost in the details: molarity used for molality, grams for kilograms, a freezing point reported as a change, or i ignored. This post works through freezing point, molar mass and association problems, and shows where the van 't Hoff factor comes from.
What makes a property colligative?
A colligative property depends on the number of solute particles in a given amount of solvent, not on their identity. The four are relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure. The JEE (Main) 2026 syllabus and the NEET (UG)-2026 syllabus list all four, along with molar mass from colligative properties, abnormal molar mass and the van 't Hoff factor.
The NCERT Class 12 chapter on Solutions gives the equations: ΔTb = iKbm, ΔTf = iKfm and π = iCRT, where m is molality. Its table gives Kf = 1.86 K kg mol⁻¹ and Kb = 0.52 K kg mol⁻¹ for water, and Kf = 5.12 K kg mol⁻¹ for benzene.
Where does the van 't Hoff factor come from?
NCERT notes that van 't Hoff introduced the factor i in 1880 to account for dissociation and association. He later won the first Nobel Prize in Chemistry, in 1901, partly for his work on osmotic pressure in solutions.
i = (number of particles after dissociation or association) ÷ (number before).
For a solute that dissociates into n ions with degree of dissociation α: i = 1 + α(n − 1).
For a solute that pairs into dimers with degree of association α: i = 1 − α/2.
NCERT's table shows that real values fall below the ideal ones except in very dilute solution. For NaCl, i is 1.87 at 0.1 m, 1.94 at 0.01 m and 1.97 at 0.001 m, against 2 for complete dissociation. For ethanoic acid in benzene, i is close to 0.5, because the molecules form dimers.
How do you solve a freezing point problem?
Example 1. Estimate the freezing point of a 0.10 mol kg⁻¹ NaCl solution in water.
Ideal i = 2: ΔTf = 2 × 1.86 × 0.10 = 0.372 K, so the freezing point is about −0.37 °C.
With NCERT's measured i of 1.87: ΔTf = 1.87 × 1.86 × 0.10 ≈ 0.348 K, about −0.35 °C.
Unless the question gives i or says to assume complete dissociation, ask yourself which it expects. Most textbook questions mean the ideal value.
How do you find molar mass?
Example 2. Dissolving 3.0 g of a non-electrolyte in 100 g of water lowers the freezing point by 0.31 K. Find its molar mass.
m = ΔTf ÷ Kf = 0.31 ÷ 1.86 ≈ 0.167 mol kg⁻¹.
Moles of solute = 0.167 × 0.100 kg ≈ 0.0167 mol.
M = 3.0 ÷ 0.0167 ≈ 180 g mol⁻¹.
In one line: M = (Kf × w₂ × 1000) ÷ (ΔTf × w₁) = (1.86 × 3.0 × 1000) ÷ (0.31 × 100) = 180 g mol⁻¹, with masses in grams. The answer matches glucose, which is a good sign.
What if the solute pairs up?
Example 3. Ethanoic acid is dissolved in benzene at 0.20 mol kg⁻¹, counted as single molecules. The freezing point falls by 0.563 K. What fraction of the acid has dimerised?
With Kf for benzene of 5.12 K kg mol⁻¹, i = ΔTf ÷ (Kf × m) = 0.563 ÷ (5.12 × 0.20) ≈ 0.55.
For dimerisation, i = 1 − α/2, so α = 2(1 − 0.55) = 0.90.
About 90% of the acid molecules are in dimers.
What do even future teachers get wrong?
Pinarbasi, Sozbilir and Canpolat (2009) gave four open-ended questions on boiling point elevation and freezing point depression to 78 prospective chemistry teachers just before they qualified, and identified nine different misconceptions. An earlier study of undergraduates by Pinarbasi and Canpolat (2003) found many could not apply solution chemistry to real-life situations. If people about to teach the topic hold such ideas, it is worth checking your own reasoning, not just your formulas.
Which slips cost the most marks?
Molarity for molality. Colligative formulas use mol per kg of solvent.
Solution mass for solvent mass. Use the solvent only.
Grams for kilograms in m.
ΔT given as the answer when the question asks for the new freezing or boiling point.
Forgetting i for electrolytes, or using the dissociation formula for a dimerising solute.
Water's Kf for another solvent.
The same particle counting reappears in ionic equilibrium; our post on pH, buffers and salt hydrolysis covers degree of dissociation from the equilibrium side.
A practice routine
Day 1: five freezing and boiling point problems, writing units on every line.
Day 2: five molar mass problems.
Day 3: four problems with dissociation or association, finding α.
Day 4: three osmotic pressure problems with π = iCRT.
Day 5: take topic-wise questions from previous-year papers and mark every slip by type.
If you keep getting the method wrong rather than the arithmetic, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Today, rework Example 3 assuming the acid had not dimerised at all, and see what freezing point drop you would predict.
