Solutions for JEE Main

Solutions is a compact chapter with a predictable set of questions: a vapour-pressure calculation with Raoult's law, a colligative-property numerical (often asking for molar mass or the degree of dissociation), and a conceptual question on deviations and azeotropes. It suits numerical-value questions well, because the arithmetic is short once you choose the right formula.

The official 2026 syllabus lists concentration terms (molality, molarity, mole fraction, percentage by mass and volume), vapour pressure and Raoult's law, ideal and non-ideal solutions with their vapour pressure–composition plots, the four colligative properties, molar mass from colligative properties, abnormal molar mass and the van 't Hoff factor. Henry's law appears in the equilibrium unit, and it fits naturally here.

Concentration terms in brief

Colligative formulas use molality for ΔTb and ΔTf and molarity for osmotic pressure. The concentration terms guide covers conversions between them.

Henry's law

The partial pressure of a gas over a solution is proportional to its mole fraction in the solution: p = K_H × x. A larger K_H means the gas is less soluble at a given pressure. Solubility of gases falls as temperature rises. Examples: soft drinks are bottled under high CO₂ pressure, and divers breathing compressed air risk the bends when dissolved N₂ comes out of the blood.

Raoult's law for two volatile liquids

For an ideal solution of A and B:

• Partial pressures: p_A = x_A P°_A and p_B = x_B P°_B
• Total pressure: P = x_A P°_A + x_B P°_B, a straight line between P°_B and P°_A
• Vapour composition (Dalton): y_A = x_A P°_A / P
• Also 1/P = y_A/P°_A + y_B/P°_B

The vapour is always richer in the more volatile component. That is the basis of fractional distillation.

Ideal and non-ideal solutions

Acetone and chloroform deviate negatively because chloroform's C–H forms a hydrogen bond with acetone's carbonyl oxygen. An azeotrope boils at constant composition, so it cannot be separated by fractional distillation.

Non-volatile solute: the four colligative properties

Colligative properties depend on the number of solute particles, not their identity. For a non-volatile solute in a dilute solution:

Rearranged for molar mass: M₂ = Kf × w₂ × 1000/(ΔTf × w₁), with masses in grams (w₁ = solvent). The same form works with Kb. Osmotic pressure is the method of choice for polymers and proteins, because even a tiny molar concentration gives a measurable π at room temperature.

Two solutions with the same osmotic pressure are isotonic. Applying a pressure larger than π on the solution side drives solvent the other way: reverse osmosis, used in desalination.

The van 't Hoff factor

i = (observed colligative property)/(calculated for no dissociation) = (normal molar mass)/(observed molar mass).

• Dissociation into n ions with degree α: i = 1 + (n − 1)α. NaCl gives up to 2, CaCl₂ 3, K₄[Fe(CN)₆] 5.
• Association of n molecules into one with degree α: i = 1 − (1 − 1/n)α. Acetic acid dimerises in benzene through hydrogen bonds, so i approaches 0.5 and the observed molar mass approaches 120.

Dissociation gives a molar mass lower than expected; association gives one higher. That is the "abnormal molar mass" in the syllabus.

Worked problems

Problem 1: vapour pressure and vapour composition. Liquids A (P° = 100 mm Hg) and B (P° = 50 mm Hg) form an ideal solution with x_A = 0.4. Find the total vapour pressure and the mole fraction of A in the vapour.

P = 0.4 × 100 + 0.6 × 50 = 40 + 30 = 70 mm Hg.
y_A = 40/70 = 0.571. The vapour is richer in A, the more volatile liquid.

Problem 2: molar mass from ΔTf. 1.8 g of a non-electrolyte in 100 g of water lowers the freezing point by 0.186 K. Find its molar mass (Kf = 1.86 K kg mol⁻¹).

M = 1.86 × 1.8 × 1000/(0.186 × 100) = 3,348/18.6 = 180 g mol⁻¹.

Problem 3: degree of dissociation. A 0.1 m aqueous solution of a salt AB₂ freezes at −0.465 °C. Find the degree of dissociation (Kf = 1.86).

Expected ΔTf with no dissociation = 1.86 × 0.1 = 0.186 K, so i = 0.465/0.186 = 2.5.
AB₂ → A²⁺ + 2B⁻ gives n = 3, so 1 + 2α = 2.5 and α = 0.75.

Problem 4 (numerical answer): relative lowering. 18 g of glucose (M = 180) is dissolved in 178.2 g of water. The vapour pressure of pure water is 24 mm Hg. Find the lowering of vapour pressure in mm Hg.

Water = 178.2/18 = 9.9 mol; glucose = 0.1 mol. x(glucose) = 0.1/10 = 0.01.
ΔP = 0.01 × 24 = 0.24.

Traps to watch

• Using molarity in ΔTf or ΔTb. They need molality: mol per kg of solvent.
• Forgetting i for electrolytes. Equal molal NaCl and glucose do not give equal effects.
• Mixing up K_H. Higher K_H means lower solubility.
• Assuming the vapour has the liquid's composition. Calculate y_A separately.
• Reversing the azeotrope types. Positive deviation, higher vapour pressure, lower boiling point, minimum-boiling azeotrope.
• Units of R in π = CRT. Use 0.0821 for atm with C in mol L⁻¹; 8.314 needs SI units.

Practice set

1. Which has the highest boiling point: 0.1 m urea, 0.1 m KCl or 0.1 m CaCl₂?
2. Find the osmotic pressure of 0.1 M glucose at 300 K (R = 0.082 L atm K⁻¹ mol⁻¹).
3. Find ΔTf for 0.1 m NaCl in water, assuming complete dissociation.
4. What kind of deviation does acetone + chloroform show, and what kind of azeotrope?
5. Find i for K₄[Fe(CN)₆], assuming complete dissociation.
6. Acetic acid is completely dimerised in benzene. Find i and the observed molar mass.
7. 6 g of urea (M = 60) is dissolved in 500 g of water. Find the boiling point (Kb = 0.52).
8. For an ideal solution, P°_A = 120 mm Hg, P°_B = 40 mm Hg and x_A = 0.25. Find y_A.

Answers

1. CaCl₂: i = 3 gives the most particles.
2. π = 0.1 × 0.082 × 300 = 2.46 atm.
3. 2 × 1.86 × 0.1 = 0.372 K.
4. Negative deviation (hydrogen bonding); maximum-boiling azeotrope.
5. 4K⁺ + [Fe(CN)₆]⁴⁻ gives i = 5.
6. i = 0.5; observed molar mass = 60/0.5 = 120 g mol⁻¹.
7. m = 0.1/0.5 = 0.2, ΔTb = 0.104 K, so bp ≈ 100.10 °C (373.25 K).
8. P = 30 + 30 = 60 mm Hg; y_A = 30/60 = 0.5.

What to do next

• Draw the vapour pressure–composition plots for ideal, positive and negative deviation from memory.
• Solve five molar-mass problems and five van 't Hoff factor problems without looking at formulas.
• Revise concentration terms, then move on to chemical kinetics.