Solutions & Colligative Properties

Solutions cover concentration measures, Raoult's law and its deviations, and the four colligative properties. Advanced emphasises the van't Hoff factor for association/dissociation and abnormal molar masses — the recurring twist that turns a routine numerical into a two-step problem.

Key results

• Raoult's law: P = x_solvent·P°; relative lowering ΔP/P° = x_solute. Positive/negative deviations signal weaker/stronger new interactions (azeotropes).
• Colligative (depend on particle number): ΔTb = i·Kb·m; ΔTf = i·Kf·m; osmotic pressure π = i·CRT.
• van't Hoff factor i: i > 1 for dissociation (i = 1 + (n−1)α), i < 1 for association (dimerisation).
• Abnormal molar mass: M_observed = M_normal/i.
• Henry's law: gas solubility p = KH·x (higher KH, lower solubility).

Property	Formula
Boiling elevation	i·Kb·m
Freezing depression	i·Kf·m
Osmotic pressure	i·CRT
Relative lowering	x_solute

Exam Tricks & Tips

• 🎯 All four colligative properties depend on the number of particles, hence on i — always compute the van't Hoff factor first for electrolytes.
• 🎯 Dissociation gives i > 1, association gives i < 1; i = 1 + (n−1)α for dissociation into n ions with degree α.
• 🎯 Abnormal (observed) molar mass = normal/i, so a dissociating solute appears lighter, an associating one heavier.
• 🎯 Osmotic pressure is the most precise colligative method for large molar masses (proteins) because π is measurable even for dilute solutions.
• 🎯 Positive deviation from Raoult's law → minimum-boiling azeotrope; negative → maximum-boiling — link deviation sign to azeotrope type.
• ❌ Common mistake: forgetting the factor i for ionic solutes — a 0.1 m NaCl solution depresses freezing point nearly twice as much as 0.1 m glucose.

Expected exam pattern

1–2 questions: a colligative-property numerical involving i (dissociation/association) and a Raoult's-law/deviation conceptual. Abnormal-molar-mass problems recur.

Quick recap

Colligative properties ∝ particle number: ΔTb = iKbm, ΔTf = iKfm, π = iCRT. i > 1 dissociation, i < 1 association; M_obs = M_normal/i. Raoult ΔP/P° = x_solute; deviation sign sets azeotrope type.

Cover the answer, recall, then check. 11 cards on solutions for JEE Advanced.

Q1. Raoult's law for the relative lowering of vapour pressure.
A1. ΔP/P° = x_solute (mole fraction of solute).

Q2. The four colligative properties.
A2. Vapour-pressure lowering, boiling-point elevation, freezing-point depression, osmotic pressure.

Q3. van't Hoff factor for dissociation into n ions with degree α.
A3. i = 1 + (n − 1)α (i > 1).

Q4. van't Hoff factor for association (e.g. dimerisation).
A4. i < 1 (fewer particles than formula units).

Q5. Boiling-point elevation with the van't Hoff factor.
A5. ΔTb = i·Kb·m.

Q6. Osmotic pressure formula.
A6. π = i·CRT.

Q7. Relation between observed and normal molar mass.
A7. M_observed = M_normal/i.

Q8. Which colligative property best measures large (polymer) molar masses?
A8. Osmotic pressure (measurable even for very dilute solutions).

Q9. Sign of Raoult's-law deviation for a minimum-boiling azeotrope.
A9. Positive deviation (weaker new interactions).

Q10. Henry's law for gas solubility.
A10. p = KH·x; higher KH means lower solubility.

Q11. Why does 0.1 m NaCl depress freezing point about twice as much as 0.1 m glucose?
A11. NaCl dissociates into two ions (i ≈ 2), doubling the particle count.

Colligative properties depend only on the number of solute particles, not their identity — which makes them a clean tool for finding molar masses and, in reverse, detecting association or dissociation. JEE Advanced tests the four properties, the van't Hoff factor, and Raoult's law with non-ideal deviations.

Core concept: dissolving a non-volatile solute lowers vapour pressure, raises boiling point, lowers freezing point and creates osmotic pressure — all proportional to solute particle concentration.

Core ideas

Beginner — Raoult's law and concentration

Raoult's law: for an ideal solution, the partial vapour pressure of each component = mole fraction × its pure vapour pressure. For a non-volatile solute, relative lowering of vapour pressure (P° − P)/P° = x_solute. Concentration terms: molarity, molality (preferred here, temperature-independent), mole fraction, ppm.

Intermediate — the four colligative properties

• Relative VP lowering: (P° − P)/P° = x_solute.
• Boiling-point elevation: ΔT_b = K_b·m (K_b = ebullioscopic constant).
• Freezing-point depression: ΔT_f = K_f·m (K_f = cryoscopic constant) — why salt melts ice.
• Osmotic pressure: π = CRT (C = molarity); the most sensitive, ideal for macromolecule molar masses.
  All are proportional to particle molality/molarity, so each can yield the solute's molar mass.

Advanced — van't Hoff factor and non-ideal solutions

Electrolytes dissociate, increasing particle count; associating solutes (dimers) decrease it. The van't Hoff factor i = (observed particles)/(formula-unit particles) multiplies every colligative expression (ΔT_f = i·K_f·m, etc.). For dissociation into n ions with degree α: i = 1 + (n−1)α; for association of n monomers: i = 1 − (1−1/n)α... i.e. i = 1 + (1/n − 1)α. Non-ideal solutions: positive deviation from Raoult's law (weaker A–B forces, e.g. ethanol + water) shows higher VP; negative deviation (stronger A–B, e.g. HNO₃ + water) shows lower VP and forms maximum-boiling azeotropes.

Worked example

0.6 g of a solute (molar mass 60) is dissolved in 100 g water. Find the freezing-point depression (K_f = 1.86 K kg⁻¹ mol⁻¹, solute non-electrolyte). Moles solute = 0.6/60 = 0.01; molality m = 0.01/0.1 kg = 0.1 mol/kg. ΔT_f = K_f·m = 1.86 × 0.1 = 0.186 K, so the solution freezes at −0.186 °C. If instead the solute were NaCl (i = 2), the depression would double to 0.372 K — the van't Hoff factor is the twist JEE adds to catch students who ignore dissociation.

How JEE Advanced tests this

Molar-mass determination from any colligative property; van't Hoff factor to find degree of dissociation/association (abnormal molar mass); Raoult's-law vapour-pressure of ideal binary mixtures and identifying positive/negative deviation; osmotic-pressure and isotonic-solution comparisons; and azeotrope reasoning.

Exam tricks & shortcuts

• Colligative ∝ particle count — multiply by i for electrolytes/associating solutes.
• Osmotic pressure π = CRT is best for large molar masses (most sensitive).
• Abnormal molar mass: observed < true ⇒ dissociation (i>1); observed > true ⇒ association (i<1).
• Mnemonic: "Count particles, not identities."

Key formulas

• Raoult: p_A = x_A p°_A; relative lowering (p°−p)/p° = x_solute.
• Henry: p = K_H x.
• Boiling point elevation: ΔT_b = i K_b m; freezing point depression ΔT_f = i K_f m.
• Osmotic pressure: π = i CRT.
• van't Hoff factor: i = observed/normal; dissociation α = (i−1)/(n−1).

Colligative properties give molar mass; the van't Hoff factor corrects for dissociation or association.

Worked Example

Problem: 0.50 g of a weak monobasic acid HA (molar mass 100 g/mol) is dissolved in 100 g of water and shows a freezing-point depression of 0.15 K. Find the degree of dissociation of the acid. (K_f for water = 1.86 K·kg·mol⁻¹.)

Solution:
Moles of acid = 0.50/100 = 0.005 mol; solvent = 0.100 kg.
Molality m = 0.005/0.100 = 0.05 mol/kg.

Expected ΔT_f if no dissociation (i = 1):
ΔT_f(calc) = K_f × m = 1.86 × 0.05 = 0.093 K.

The van 't Hoff factor from the observed depression:
i = ΔT_f(observed)/ΔT_f(calc) = 0.15/0.093 = 1.61.

For a monobasic acid HA ⇌ H⁺ + A⁻, one particle becomes (1 + α):
i = 1 + α → α = i − 1 = 1.61 − 1 = 0.61.

Answer: Degree of dissociation α ≈ 0.61 (about 61%).