Chapters

Everything NCERT Class 12 Unit 1 (Solutions) says that can be asked, plus what JEE Main adds beyond the book (marked Beyond NCERT). The short note stays for the last week; this is for a proper revision pass.

Formula sheet

Every formula in the chapter on one screen, for the last pass. Details and examples are in the sections below.

Quantity	Formula	Watch
Mass %	solute / solution × 100	Also V/V and w/V (g per 100 mL)
ppm	part / whole × 10⁶	By mass unless told
Mole fraction	xᵢ = nᵢ / Σn	Σx = 1
Molarity	mol solute / L solution	Changes with T
Molality	mol solute / kg solvent	Solvent, not solution
Henry's law	p = KH x	Bigger KH, less soluble
Raoult, two liquids	p = x₁p₁° + x₂p₂°	Linear in x₂
Vapour composition	yᵢ = xᵢpᵢ° / p	Vapour richer in the more volatile
Relative lowering	(p₁° − p₁)/p₁° = x₂ ≈ n₂/n₁	× i for electrolytes
Boiling point rise	ΔTb = i Kb m	m = molality
Freezing point fall	ΔTf = i Kf m	m = molality
Molar mass from ΔT	M₂ = 1000 K w₂ / (ΔT w₁)	w in g
K from enthalpy	Kf = R M₁ Tf² / (1000 ΔfusH)	M₁ in g mol⁻¹
Osmotic pressure	π = i C R T	C = molarity
van 't Hoff factor	i = normal M / observed M	i > 1 dissociation, < 1 association
Dissociation	i = 1 + (n − 1)α	n ions per formula
Association	i = 1 − α(1 − 1/n)	n molecules → 1

0 · What JEE Main asks from this chapter

202 questions in JEE Main 2019–2026, 1 to 2 per paper, mostly easy numericals. Grouped by type (counts are approximate):

Type	Share	Typical question
Concentration: M, m, x, %, ppm, conversions with density	~25%	0.8 M H₂SO₄, density 1.06 → molality
ΔTf and ΔTb: freezing/boiling point, molar mass, ice separated	~22%	62 g glycol in 250 g water at −10 °C → ice formed
van 't Hoff factor: α, association, Kₐ from ΔTf, comparing solutions	~21%	A₂B₃, 60% ionised, 1 m → boiling point
Raoult's law, ideal and non-ideal solutions, azeotropes	~15%	Equimolar benzene + toluene → mole fraction in vapour
Osmotic pressure, isotonic solutions, reverse osmosis	~12%	Blood at 7.47 bar → isotonic glucose in g L⁻¹
Henry's law	~5%	CO₂ at 0.835 bar, KH 1.67 kbar → mmol dissolved

Most numericals are rounded to the nearest integer, often as "x × 10⁻²". Read the multiplier before you answer.

1 · Types of solution

• Solution: a homogeneous mixture of two or more components; composition and properties uniform throughout.
• Solvent: usually the component present in the largest amount; it decides the physical state of the solution. The others are solutes. This unit takes only binary solutions.

Type	Solute in solvent	Example
Gaseous	gas in gas	O₂ + N₂
	liquid in gas	chloroform in N₂
	solid in gas	camphor in N₂
Liquid	gas in liquid	O₂ in water
	liquid in liquid	ethanol in water
	solid in liquid	glucose in water
Solid	gas in solid	H₂ in palladium
	liquid in solid	sodium amalgam (Hg in Na)
	solid in solid	copper in gold

Alloys from NCERT's opening: brass is Cu + Zn, bronze is Cu + Sn, German silver is Cu + Zn + Ni.

2 · Expressing concentration

The units in detail, with conversions, are in Mole Concept §9. What this unit adds:

1. Mass % (w/w) = mass of component / mass of solution × 100. 10% glucose = 10 g glucose + 90 g water.
2. Volume % (V/V) = volume of component / volume of solution × 100. 35% (V/V) ethylene glycol is car antifreeze; it lowers the freezing point of water to 255.4 K (−17.6 °C).
3. Mass by volume % (w/V) = grams of solute in 100 mL of solution. Used in medicine and pharmacy (normal saline is 0.9% w/V).
4. ppm = parts of component / total parts × 10⁶, as mass/mass, volume/volume or mass/volume. Sea water: 6 × 10⁻³ g O₂ in 1030 g = 5.8 ppm.
5. Mole fraction, molarity, molality as before.

Temperature: mass %, ppm, mole fraction and molality don't change with temperature; molarity does, because volume changes. (2024: "Which quantity changes with temperature?" Molarity.)

d is in g mL⁻¹, M₂ is the solute's molar mass, M₁ the solvent's.

Example (NCERT 1.1) · 20% ethylene glycol by mass. Take 100 g: 20/62 = 0.322 mol glycol, 80/18 = 4.444 mol water. x(glycol) = 0.322/4.766 = 0.068; x(water) = 0.932.

Example (2024) · 0.8 M H₂SO₄, density 1.06 g cm⁻³. 1 L weighs 1060 g and holds 0.8 × 98 = 78.4 g acid, so water = 981.6 g. m = 0.8/0.9816 = 0.815 m (answer 815 as × 10⁻³).

Example (2021) · Mole fraction of solute in a 100 molal aqueous solution: 100 mol solute per 1000/18 = 55.6 mol water → x = 100/155.6 = 0.64 (answer 64 as × 10⁻²).

3 · Solubility

Solubility is the maximum amount that dissolves in a given amount of solvent at a given temperature. It depends on the nature of solute and solvent, temperature and pressure.

Like dissolves like: polar solutes dissolve in polar solvents, non-polar in non-polar. NaCl and sugar dissolve in water, not benzene; naphthalene and anthracene dissolve in benzene, not water.

• Dissolution: solute going into solution. Crystallisation: solute particles coming back out. When the two rates are equal the solution is in dynamic equilibrium: solute + solvent ⇌ solution.
• Saturated solution: no more solute dissolves at that T and P; it is in equilibrium with undissolved solute, and its concentration is the solubility. Unsaturated: more can dissolve.

Solid in a liquid:

• Temperature (Le Chatelier): if ΔsolH > 0 (endothermic), solubility rises with temperature; if ΔsolH < 0, it falls.
• Pressure: no significant effect; solids and liquids are nearly incompressible.

Gas in a liquid: solubility rises with pressure and falls with temperature.

Henry's law: at constant temperature, the solubility of a gas in a liquid is proportional to the partial pressure of the gas above it. The common form:

• p = partial pressure of the gas in the vapour; x = its mole fraction in the solution. A plot of p against x is a straight line through the origin with slope KH.
• KH depends on the gas (and the solvent) and on temperature. Higher KH at a given pressure means lower solubility.
• KH rises with temperature (N₂: 76.48 kbar at 293 K, 88.84 at 303 K), so gases are less soluble in warm water. Aquatic life is more comfortable in cold water.
• Dissolving a gas is like condensation, so it is exothermic; by Le Chatelier, solubility falls as T rises.

Example (NCERT 1.4) · N₂ at 0.987 bar, KH = 76.48 kbar at 293 K. x = 0.987/76 480 = 1.29 × 10⁻⁵. 1 L water = 55.5 mol, so n = 1.29 × 10⁻⁵ × 55.5 = 7.16 × 10⁻⁴ mol = 0.716 mmol.

Example (2021) · CO₂ at 0.835 bar, KH = 1.67 × 10³ bar. x = 5.0 × 10⁻⁴. 0.9 L water = 50 mol → n = 0.025 mol = 25 mmol.

Applications of Henry's law:

• Soft drinks and soda water are sealed under high CO₂ pressure to raise its solubility.
• Bends: scuba divers breathe air at high pressure, so more N₂ dissolves in the blood; surfacing releases it as bubbles that block capillaries. Their tanks hold air diluted with helium (11.7% He, 56.2% N₂, 32.1% O₂).
• Anoxia: at high altitude the partial pressure of O₂ is low, so blood O₂ is low; climbers become weak and unable to think clearly.

JEE asks (2019): "Which statement about Henry's law is not correct?" Answer: "higher KH, higher solubility". It is the reverse.

4 · Vapour pressure of liquid solutions

Raoult's law (two volatile liquids): the partial vapour pressure of each component is proportional to its mole fraction in the solution.

• p° is the vapour pressure of the pure component at the same temperature. Dalton's law gives p(total) = p₁ + p₂.
• p(total) varies linearly with x₂, between p₁° and p₂°.
• Vapour composition by Dalton's law: pᵢ = yᵢ p(total). The vapour is always richer in the more volatile component (higher p°).

The second form (Beyond NCERT) gives p(total) straight from the vapour composition.

Example (NCERT 1.5) · 25.5 g CHCl₃ (0.213 mol, p° = 200 mm Hg) + 40 g CH₂Cl₂ (0.47 mol, p° = 415). x(CH₂Cl₂) = 0.688. p = 200 + 215 × 0.688 = 347.9 mm Hg. y(CH₂Cl₂) = 0.688 × 415 / 347.9 = 0.82.

Example (2021) · Equimolar benzene (70 torr) and toluene (20 torr): p = 35 + 10 = 45 torr; y(benzene) = 35/45 = 0.78.

Example (2020, two mixtures) · 1 mol hexane + 3 mol heptane: 550 mm Hg; one more mole of heptane: 560. So p(hex) + 3p(hep) = 2200 and p(hex) + 4p(hep) = 2800 → p°(heptane) = 600 mm Hg (hexane 400). The same trick came back in 2023, 2025 and 2026.

Raoult's law as a special case of Henry's law. Both say partial pressure ∝ mole fraction; only the constant differs. Raoult's law is Henry's law with KH = p₁°.

Solids in liquids: with a non-volatile solute only the solvent is in the vapour. Solute particles take part of the surface, fewer solvent molecules escape, and the vapour pressure falls. The fall depends on the amount of solute, not its nature: 1 mol sucrose or 1 mol urea in 1 kg water lower it about equally.

General Raoult's law: for any solution, the partial vapour pressure of each volatile component is proportional to its mole fraction.

5 · Ideal and non-ideal solutions

Ideal solution: obeys Raoult's law over the entire range of concentration. Also ΔmixH = 0 and ΔmixV = 0: no heat change on mixing, and the volumes simply add. A–B attractions are nearly equal to A–A and B–B. Nearly ideal: n-hexane + n-heptane, bromoethane + chloroethane, benzene + toluene.

Non-ideal solution: doesn't obey Raoult's law over the whole range; its vapour pressure is higher or lower than Raoult predicts.

	Positive deviation	Negative deviation
A–B attraction	weaker than A–A, B–B	stronger than A–A, B–B
Vapour pressure	higher than Raoult	lower than Raoult
ΔmixH, ΔmixV · Beyond NCERT	> 0 (absorbs heat, expands)	< 0 (releases heat, contracts)
Boiling point	lower	higher
Azeotrope (if large)	minimum boiling	maximum boiling
NCERT examples	ethanol + acetone (acetone breaks ethanol's H-bonds); CS₂ + acetone	phenol + aniline (O–H···N bond); chloroform + acetone (C–H···O bond)

The straight line is Raoult's law for the total; the faint dashed lines are the partial pressures p₁ and p₂. A curve above the line is positive deviation, below is negative.

Example (2020) · CS₂ (512 mm Hg) and acetone (144) give a total of 600 mm Hg, above anything Raoult allows (at most 512). So positive deviation: they attract each other less than themselves, heat is absorbed, and 100 mL + 100 mL gives more than 200 mL. The false statement was "volume < 200 mL".

Azeotropes: binary mixtures with the same composition in liquid and vapour, boiling at a constant temperature. Fractional distillation can't separate them.

• Minimum boiling azeotrope: from a large positive deviation. Ethanol–water: fractional distillation stops at about 95% ethanol by volume.
• Maximum boiling azeotrope: from a large negative deviation. Nitric acid–water: about 68% HNO₃, 32% water by mass, boiling at 393.5 K.

Example (2025) · Which does not form a minimum boiling azeotrope: CH₃OH + CHCl₃, CS₂ + acetone, phenol + aniline, water + butanone? Phenol + aniline: it shows negative deviation.

6 · Colligative properties

Four properties of a solution of a non-volatile solute that depend only on the number of solute particles, not their identity (Latin co, together; ligare, to bind):

1. relative lowering of vapour pressure
2. elevation of boiling point
3. depression of freezing point
4. osmotic pressure

6.1 Relative lowering of vapour pressure equals the mole fraction of the solute (Raoult). With several solutes it depends on the sum of their mole fractions.

The approximation is for dilute solutions (n₂ ≪ n₁). If the question says "solution is not dilute", keep n₂ in the denominator.

Example (NCERT 1.6) · Benzene p° = 0.850 bar, solution 0.845 bar, 0.5 g solid in 39.0 g benzene. 0.005/0.850 = (0.5 × 78)/(M₂ × 39) → M₂ = 170 g mol⁻¹.

Example (2019) · 0.60 g urea (0.01 mol) in 360 g water (20 mol), p° = 35 mm Hg. Δp = 35 × 0.01/20.01 = 0.017 mm Hg.

6.2 Elevation of boiling point. A liquid boils when its vapour pressure equals the atmospheric pressure (water: 1.013 bar at 373.15 K). The solution's vapour pressure is lower, so it must be heated more. 1 mol sucrose in 1000 g water boils at 373.52 K.

Kb is the molal elevation (ebullioscopic) constant, unit K kg mol⁻¹. Water: 0.52.

Example (NCERT 1.7, 1.8) · 18 g glucose (0.1 mol) in 1 kg water: ΔTb = 0.052 K → boils at 373.202 K. 1.80 g solute in 90 g benzene raises the b.p. from 353.23 K to 354.11 K: M₂ = 2.53 × 1.8 × 1000/(0.88 × 90) = 58 g mol⁻¹.

6.3 Depression of freezing point. The freezing point is the temperature at which the liquid's vapour pressure equals the solid's. The solution's lower vapour pressure meets the solid's curve at a lower temperature. Only the solvent freezes out (2023).

Kf is the molal depression (cryoscopic) constant, K kg mol⁻¹. Water: 1.86.

The solution's curve (dashed) lies below the pure solvent's everywhere. It reaches 1 atm at a higher temperature (ΔTb) and meets the solid's curve at a lower one (ΔTf).

Both constants come from the solvent alone:

M₁ in g mol⁻¹, T in kelvin. Since ΔfusH = Tf ΔfusS, Kf can also be written R M₁ Tf /(1000 ΔfusS) (asked 2025).

Example (NCERT 1.9, 1.10) · 45 g ethylene glycol in 600 g water: 0.73 mol/0.6 kg = 1.2 m → ΔTf = 2.2 K, f.p. 270.95 K. 1.00 g solute in 50 g benzene lowers its f.p. by 0.40 K: M₂ = 5.12 × 1 × 1000/(0.40 × 50) = 256 g mol⁻¹.

Example (2019, ice separated) · 62 g glycol (1 mol) in 250 g water, cooled to −10 °C. Ice forms until the remaining water makes m = 10/1.86 = 5.38 m: water left = 1/5.38 = 0.186 kg. Ice = 250 − 186 = 64 g.

Example (2023) · Solvents X and Y, same molar mass, b.p. 2 : 1, ΔvapH 1 : 2. Kb ∝ Tb²/ΔvapH → (4/1) × (2/1) = 8.

6.4 Osmosis and osmotic pressure.

• Semipermeable membrane (SPM): lets small solvent molecules through but not solute. Natural (pig's bladder, parchment) or synthetic (cellophane).
• Osmosis: flow of solvent through an SPM from pure solvent (or dilute solution) into the more concentrated solution.
• Osmotic pressure (π): the extra pressure on the solution that just stops osmosis.

C is molarity (V in litres). Use R = 0.083 L bar mol⁻¹ K⁻¹ for bar, 0.0821 L atm for atm, 8.314 with SI units (π in Pa, V in m³).

Why osmotic pressure suits proteins and polymers: it is measured at room temperature (biomolecules decompose on heating), uses molarity, and is large even for very dilute solutions (polymers dissolve poorly).

Example (NCERT 1.11) · 1.26 g protein in 200 cm³, π = 2.57 × 10⁻³ bar at 300 K. M₂ = 1.26 × 0.083 × 300/(2.57 × 10⁻³ × 0.200) ≈ 61 000 g mol⁻¹.

Isotonic solutions have the same osmotic pressure at the same temperature; no osmosis between them.

• Blood cell fluid is isotonic with 0.9% (w/V) NaCl, normal saline, safe to inject.
• Hypertonic (more than 0.9%): water leaves the cell, it shrinks. Hypotonic (less): water enters, it swells.
• Everyday osmosis: raw mango shrivels in brine; wilted flowers and limp carrots revive in water; salty food causes edema; water rises from soil into plants; salt and sugar preserve meat and fruit by killing bacteria through water loss.

Example (2022) · Blood π = 7.47 bar at 300 K. C = 7.47/(0.083 × 300) = 0.30 M → glucose 0.30 × 180 = 54 g L⁻¹.

Example (2020, mixing) · 1 L NaCl solution (π = 0.10 atm) + 2 L glucose (π = 0.20 atm). π adds by particles: (1 × 0.10 + 2 × 0.20)/3 = 0.167 atm.

Osmotic height · Beyond NCERT · When the rise of solution in a tube balances osmosis, π = hρg (2026 used an 80 mm column to find the molar mass of haemoglobin).

Reverse osmosis: a pressure larger than π on the solution side pushes pure solvent out of the solution through the SPM. Used to desalinate sea water; the membrane is cellulose acetate on a support (permeable to water, not to ions and impurities).

7 · Abnormal molar masses and the van 't Hoff factor

Colligative properties count particles, so a solute that dissociates or associates gives a "wrong" molar mass.

• Dissociation (KCl → K⁺ + Cl⁻): more particles, bigger effect, observed molar mass lower than true. 1 mol KCl in 1 kg water should raise the b.p. by 2 × 0.52 = 1.04 K.
• Association (2 CH₃COOH ⇌ (CH₃COOH)₂ in benzene, by H-bonding, in solvents of low dielectric constant): fewer particles, observed molar mass higher. Full dimerisation halves ΔT and doubles the apparent molar mass.

i > 1 for dissociation (KCl ≈ 2), i < 1 for association (acetic acid in benzene ≈ 0.5). Every formula takes i:

i from α · Beyond NCERT (NCERT only works the two cases below):

Solute	n (ions)	i at full dissociation
NaCl, KCl, MgSO₄	2	2
K₂SO₄, CaCl₂, MgCl₂	3	3
AlCl₃, K₃[Fe(CN)₆]	4	4
Al₂(SO₄)₃, K₄[Fe(CN)₆]	5	5

NCERT Table 1.4: real i is below the ideal value and rises towards it on dilution. NaCl: 1.87 (0.1 m), 1.94 (0.01 m), 1.97 (0.001 m). MgSO₄: 1.21, 1.53, 1.82. K₂SO₄: 2.32, 2.70, 2.84. So i(0.1 M) < i(0.01 M) < i(0.001 M) (asked 2024).

Example (NCERT 1.12) · 2 g benzoic acid in 25 g benzene, ΔTf = 1.62 K, Kf = 4.9. Observed M = 4.9 × 2 × 1000/(25 × 1.62) = 241.98. i = 122/241.98 = 0.504 = 1 − x/2 → x = 0.992, 99.2% associated.

Example (NCERT 1.13) · 0.6 mL acetic acid (d 1.06) = 0.0106 mol in 1 L water. Calculated ΔTf = 1.86 × 0.0106 = 0.0197 K; observed 0.0205 K. i = 1.041 = 1 + α → α = 0.041. Kₐ = cα²/(1 − α) = 0.0106 × 0.041²/0.959 = 1.86 × 10⁻⁵.

Example (2021) · 1 m A₂B₃, 60% ionised: i = 1 + 4 × 0.6 = 3.4. ΔTb = 3.4 × 0.52 = 1.77 K → 375 K.

Example (2025) · MX₂: observed M 65.6, normal 164 → i = 2.5 = 1 + 2α → α = 75%.

Example (2021, both at once) · In a solvent, 50% of HA dimerises and the rest dissociates. Particles per mole: 0.5/2 + 0.5 × 2 = 1.25.

Comparing solutions: rank by i × concentration. Bigger i × m → lower f.p., higher b.p., higher π, lower vapour pressure.

Example (2021) · Lowest freezing point among 0.06 M Al₂(SO₄)₃, glucose, KI, K₂SO₄: i × C = 0.30, 0.06, 0.12, 0.18 → Al₂(SO₄)₃.

8 · Traps

Trap	What goes wrong	Do this
Henry's KH	Thinking big KH means more soluble	p = KH x: big KH, small x
ΔT vs π	Using molarity for ΔT or molality for π	ΔT: kg of solvent; π: L of solution
Forgetting i	Treating NaCl or K₄[Fe(CN)₆] as one particle	Count ions; K₄[Fe(CN)₆] gives 5
Association	Using i = 1 for acids in benzene	Carboxylic acids dimerise in benzene: i ≈ ½
Vapour vs liquid	Using x where y is asked	y = xp°/p; vapour has more of the volatile one
Kelvin vs Celsius	Mixing them in boiling or freezing points	ΔT is the same in K and °C; T in π and Kf must be K
Non-dilute RLVP	Using n₂/n₁ when the question says "not dilute"	Keep n₁ + n₂ (or n₁ + i n₂) below
Ice separation	Reporting water left instead of ice	Ice = water at start − water left
Deviation signs	Mixing up ΔV for positive deviation	Positive: weaker A–B, ΔH > 0, ΔV > 0, min-boiling
Volatile solute (2026)	Assuming it always raises the b.p.	Only a non-volatile solute raises the b.p.
Units of R	Using 0.083 with atm	0.083 L bar, 0.0821 L atm, 8.314 J (SI)

9 · Numbers worth knowing

Solvent	Kb (K kg mol⁻¹)	Kf (K kg mol⁻¹)
Water	0.52	1.86
Benzene	2.53	5.12
Ethanol	1.20	1.99
Chloroform	3.63	4.79
CCl₄	5.03	31.8
Acetic acid	2.93	3.90
Cyclohexane	2.79	20.0

• Water: Kf > Kb. Benzene: Kf 5.12 > water's 1.86.
• 1 L water = 55.5 mol. Normal saline = 0.9% (w/V) NaCl.
• Ethanol–water azeotrope ≈ 95% (V/V), minimum boiling. HNO₃–water ≈ 68% (by mass), maximum boiling at 393.5 K.
• KH (kbar, in water): CO₂ 1.67 (298 K), O₂ 34.86 (293 K), N₂ 76.48 (293 K), He 144.97 (293 K).
• R = 0.083 L bar mol⁻¹ K⁻¹ = 0.0821 L atm = 8.314 J.

10 · For the boards only

JEE Main doesn't ask this, but the CBSE paper can.

• 1 ppm fluoride in water prevents tooth decay; 1.5 ppm mottles teeth; high concentrations are poisonous (sodium fluoride is used in rat poison).
• Commercial bleach: 3.62% (by mass) sodium hypochlorite.
• Intravenous injections are made in water with the same ionic concentrations as blood plasma.
• Henry was the first to give the pressure–solubility law; Dalton reached the same idea independently. Raoult stated his law in 1886. van 't Hoff introduced i in 1880.
• Desalination plants supply potable water in many countries by reverse osmosis.