IIT-JAM Solutions and Colligative Properties — Full Numerical Method
Solutions is a unit an IIT-JAM aspirant should treat as guaranteed marks. The physics is shallow, the formulas are few, and the arithmetic is short — yet students lose marks here more often than in quantum chemistry, almost always for the same two reasons: mixing up molality with molarity, and forgetting the van't Hoff factor. This article covers the whole unit at JAM depth, with every worked example computed line by line.
Concentration units — get these right first
Molality m = moles of solute ÷ mass of solvent in kilograms (mol/kg)
Mole fraction xA = nA ÷ (nA + nB), and xA + xB = 1
Molality and mole fraction depend only on masses, so they do not change with temperature. Molarity does, because the volume of the solution expands on heating. That is exactly why the colligative-property formulas for freezing and boiling points are written in molality, while osmotic pressure — measured at a fixed temperature — is written in molarity.
Raoult's law and the ideal solution
Vapour composition: yA = pA ÷ ptotal
Here p°A is the vapour pressure of pure A, x is the mole fraction in the liquid and y in the vapour. An ideal solution obeys this over the whole composition range, has ΔHmix = 0 and ΔVmix = 0.
Example 1. Two volatile liquids form an ideal solution. At the working temperature p°A = 100 kPa and p°B = 40 kPa. Mix 2.0 mol of A with 3.0 mol of B. Find the total pressure and the vapour composition.
Total moles = 2.0 + 3.0 = 5.0 → xA = 2.0/5.0 = 0.40, xB = 0.60.
pA = 0.40 × 100 = 40.0 kPa
pB = 0.60 × 40 = 24.0 kPa
ptotal = 40.0 + 24.0 = 64.0 kPa
yA = 40.0 ÷ 64.0 = 0.625, yB = 1 − 0.625 =
0.375.
Notice yA (0.625) > xA (0.40): the vapour is always richer in the more volatile component. That single sentence is the basis of fractional distillation, and JAM asks it as a reasoning question.
Non-ideal solutions. If A–B interactions are weaker than the A–A and B–B interactions, molecules escape more easily, the observed pressure exceeds Raoult's prediction (positive deviation, ΔHmix > 0), and the mixture can form a minimum-boiling azeotrope. If A–B interactions are stronger, you get negative deviation, ΔHmix < 0 and a maximum-boiling azeotrope. For a dilute solution the solvent follows Raoult's law and the solute follows Henry's law, p = KHx, which is the same straight line with a different slope.
The four colligative properties
Colligative means the effect depends on the number of solute particles, not their identity. All four formulas below carry the van't Hoff factor i.
2. Boiling point elevation: ΔTb = i·Kb·m
3. Freezing point depression: ΔTf = i·Kf·m
4. Osmotic pressure: π = i·M·R·T
Kb and Kf are the ebullioscopic and cryoscopic constants of the solvent (units K kg mol−1) and are always supplied in the question. For osmotic pressure, M is molarity in mol/L, T is in kelvin and R = 0.083145 L bar K−1 mol−1 when you want π in bar.
Example 2 — freezing point of a glucose solution. Dissolve 1.80 g of glucose (C₆H₁₂O₆) in 100 g of water. Kf(water) = 1.86 K kg mol−1, Kb(water) = 0.512 K kg mol−1. Find the freezing and boiling points.
Molar mass of glucose = 6(12.011) + 12(1.008) + 6(15.999)
= 72.066 + 12.096 + 95.994 = 180.156 g/mol.
Moles = 1.80 ÷ 180.156 = 0.009991 mol.
Mass of solvent = 100 g = 0.100 kg.
Molality m = 0.009991 ÷ 0.100 = 0.09991 mol/kg.
Glucose is a non-electrolyte, so i = 1.
ΔTf = 1 × 1.86 × 0.09991 = 0.18584 → 0.186 K, so the solution freezes
at −0.186 °C.
ΔTb = 1 × 0.512 × 0.09991 = 0.05115 → 0.051 K, so it boils at
100.051 °C.
Sanity check: Kf is about 3.6 times Kb for water, and 0.186 ÷ 0.051 = 3.6. The two answers are consistent.
The van't Hoff factor — where most marks are lost
Dissociation into n particles: i = 1 + (n − 1)α
Association into an n-mer: i = 1 − α(1 − 1/n)
α is the degree of dissociation or association. Dissociation gives i > 1 (more particles), association gives i < 1 (fewer particles). For complete dissociation put α = 1: NaCl gives i = 2, K₂SO₄ gives i = 3, AlCl₃ gives i = 4.
Example 3 — finding α from a measured freezing point. A 0.100 m aqueous solution of K₂SO₄ freezes 0.465 K below pure water. Kf = 1.86 K kg mol−1. Find i and α.
Expected depression if there were no dissociation:
ΔTf(calc) = 1.86 × 0.100 = 0.186 K.
i = 0.465 ÷ 0.186 = 2.50.
K₂SO₄ → 2K⁺ + SO₄²⁻, so n = 3.
i = 1 + (n − 1)α → 2.50 = 1 + 2α → 2α = 1.50 → α = 0.75, i.e. 75 % dissociated.
Check by counting particles: from 1 mol of K₂SO₄, 0.75 mol dissociates giving 2.25 mol of ions, and 0.25 mol stays intact. Total = 2.25 + 0.25 = 2.50 mol of particles per mole taken — which is i, exactly as computed.
Example 4 — molar mass of a macromolecule from osmotic pressure. 1.00 g of a protein in 100 mL of aqueous solution shows an osmotic pressure of 2.50 × 10−3 bar at 300 K. Find the molar mass.
π = (w/M)(1/V)RT, so M = wRT ÷ (πV).
Numerator: 1.00 × 0.083145 × 300 = 24.9435 g L bar mol−1.
Denominator: (2.50 × 10−3 bar) × (0.100 L) = 2.50 × 10−4 L bar.
M = 24.9435 ÷ (2.50 × 10−4) = 99 774 ≈ 9.98 × 104 g/mol.
Why osmotic pressure and not freezing point for a protein? Put the same solution through Example 2's method: the molality is about 1.0 × 10−4 mol/kg, giving ΔTf ≈ 1.86 × 1.0 × 10−4 ≈ 1.9 × 10−4 K — far too small to measure. Osmotic pressure is the only one of the four that is large enough for macromolecules, and JAM expects that reasoning.
Example 5 — molar mass from relative lowering. 5.0 g of a non-volatile, non-electrolytic solute in 95 g of water lowers the vapour pressure from 100.0 kPa to 98.0 kPa. Find the molar mass of the solute.
(p° − p) ÷ p° = (100.0 − 98.0) ÷ 100.0 = 0.0200 = xsolute.
Moles of water = 95 ÷ 18.015 = 5.2734 mol.
xsolute = ns ÷ (ns + 5.2734) = 0.0200
→ ns = 0.0200 ns + 0.105468
→ 0.9800 ns = 0.105468 → ns = 0.10762 mol.
M = 5.0 ÷ 0.10762 = 46.5 g/mol.
Common mistakes that cost marks
- Dividing by the mass of the solution instead of the solvent when finding molality. In Example 2 the divisor is 0.100 kg of water, not 0.1018 kg of solution.
- Leaving out i for an electrolyte. If the solute is ionic and the question does not say "assume no dissociation", i is not 1.
- Using i > 1 for an associating solute. Carboxylic acids dimerise in benzene, so i < 1 there — for complete dimerisation i = 0.5.
- Mixing Kb and Kf. They belong to different properties and differ by a factor of about 3.6 for water; swapping them is an instant wrong answer that still "looks" reasonable.
- Forgetting to convert temperature to kelvin in π = iMRT, and forgetting that ΔT is a difference, so a value in K equals the same value in °C.
- Using molality in the osmotic-pressure formula. π needs molarity — the volume of solution.
- Assuming the vapour and liquid have the same composition. They only do at an azeotrope or for a pure liquid.
How this unit is examined
| Question shape | What it actually tests | Route to the answer |
|---|---|---|
| Find the freezing/boiling point of a given solution | Molality and i | moles → molality → ΔT = iKm |
| Find molar mass from a colligative measurement | Rearranging correctly | Solve the same formula for M |
| Find α from an observed ΔT | van't Hoff factor | i = observed ÷ calculated, then i = 1 + (n − 1)α |
| Rank solutions by boiling point | Particle counting | Compare i × m, not m alone |
| Vapour composition of an ideal mixture | Raoult's law | pA = xAp°A, then y = pA/ptotal |
| Positive or negative deviation, azeotrope type | Intermolecular reasoning | Compare A–B with A–A and B–B forces |
The exact syllabus wording, question count and marking scheme change from year to year — take those from the current official IIT-JAM notification, never from a summary page.
Check every concentration before you use it. Most lost marks in this unit are a concentration error, not a formula error. The free Concentration Calculator converts between mass, moles, volume and molarity so you can confirm the number you are about to feed into π = iMRT or ΔT = iKm.
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