Solutions & Colligative Properties

Concept Core

Concentration, Raoult's law, colligative properties, and van't Hoff factor.

Concentration Units — Quick Reference

Unit	Definition	Temperature Dependent?
Molarity (M)	moles of solute per litre of solution	Yes (volume changes with T)
Molality (m)	moles of solute per kg of solvent	No
Mole fraction (χ)	moles of component / total moles	No
Mass %	(mass of solute / mass of solution) × 100	No

Preferred for colligative properties: Molality (m) — temperature independent.

Raoult's Law & Vapour Pressure

For an ideal solution, partial vapour pressure of a component = mole fraction × pure vapour pressure:

P_A = χ_A × P°_A P_total = P_A + P_B Relative lowering of VP: ΔP/P°_A = χ_B (mole fraction of solute) = n_B/(n_A + n_B) ≈ n_B/n_A (dilute solution)

Elevation of Boiling Point

ΔT_b = K_b × m = K_b × (w_B/M_B) × (1000/w_A) where K_b = molal elevation constant (ebullioscopic constant) w_B = mass of solute (g); M_B = molar mass of solute w_A = mass of solvent (g)

K_b for water = 0.52 K·kg/mol. Boiling point increases with solute addition.

Depression of Freezing Point

ΔT_f = K_f × m = K_f × (w_B/M_B) × (1000/w_A) where K_f = molal depression constant (cryoscopic constant)

K_f for water = 1.86 K·kg/mol (larger than K_b). Anti-freeze in cars uses this principle.

Freezing point is DEPRESSED (decreased); Boiling point is ELEVATED (increased).

Osmotic Pressure

π = CRT = (n/V)RT (van't Hoff equation) π = MRT (M = molarity) For electrolytes: π = iCRT (i = van't Hoff factor)

Osmosis: solvent moves from dilute (low π) to concentrated (high π) across a semi-permeable membrane.

van't Hoff Factor (i)

i = observed colligative property / calculated value (for non-electrolyte) For electrolytes: i = 1 + (n−1)α where n = number of ions; α = degree of dissociation NaCl: i ≈ 2 (fully dissociates → 2 ions) K₂SO₄: i ≈ 3 (→ 3 ions)

For association: i < 1 (fewer particles than expected).

Formula Vault

All solution and colligative property formulas.

Molality

m = (w_B/M_B) × (1000/w_A)

w_B in g; w_A in g; M_B in g/mol

Raoult's Law

P_A = χ_A × P°_A

For ideal solution; χ_A = mole fraction

Relative VP Lowering

ΔP/P° = χ_B

χ_B = mole fraction of solute

Boiling Point Elevation

ΔT_b = K_b × m

K_b(water) = 0.52 K·kg/mol

Freezing Point Depression

ΔT_f = K_f × m

K_f(water) = 1.86 K·kg/mol

Osmotic Pressure

π = CRT = MRT

C = molarity; T in Kelvin; R = 0.0821 L·atm/mol/K

van't Hoff Factor

i = 1 + (n−1)α

n = ions formed; α = degree of dissociation

Henry's Law

p = K_H × χ (gas)

Solubility of gas ∝ its partial pressure

Worked Examples

5 problems — molality, Raoult's law, boiling point, osmotic pressure, and van't Hoff factor.

EasyFind boiling point elevation: 10g glucose (M=180) in 100g water. K_b=0.52▾

10 g of glucose (M=180 g/mol) is dissolved in 100 g of water. K_b = 0.52 K·kg/mol. Find ΔT_b.

m = (w_B/M_B) × (1000/w_A) = (10/180) × (1000/100) = 0.0556 × 10 = 0.556 mol/kg

ΔT_b = K_b × m = 0.52 × 0.556 = 0.289 K ≈ 0.29°C

✓ ΔT_b = 0.29 K (boiling point rises from 100°C to 100.29°C)

EasyFind osmotic pressure: 0.5 M sucrose solution at 27°C▾

Calculate the osmotic pressure of 0.5 M sucrose solution at 27°C. R = 0.0821 L·atm/mol/K.

T = 300 K, C = 0.5 mol/L

π = CRT = 0.5 × 0.0821 × 300 = 12.32 atm

✓ π = 12.32 atm

MediumRelative lowering of vapour pressure: 9g glucose in 90g water. M_glucose=180, M_water=18.▾

Find the relative lowering of vapour pressure when 9 g of glucose is dissolved in 90 g of water.

n_glucose = 9/180 = 0.05 mol; n_water = 90/18 = 5 mol

χ_glucose = 0.05/(0.05+5) = 0.05/5.05 ≈ 0.00990

ΔP/P° = χ_solute = 0.00990 ≈ 0.01

✓ Relative VP lowering = 0.0099 ≈ 1%

EAPCET LevelFind molar mass of solute: 5g in 50g water, ΔT_f = 0.372°C, K_f = 1.86▾

5 g of a non-electrolyte dissolved in 50 g of water causes ΔT_f = 0.372 K. K_f(water) = 1.86 K·kg/mol. Find molar mass of solute.

ΔT_f = K_f × m → m = ΔT_f/K_f = 0.372/1.86 = 0.2 mol/kg

m = (w_B/M_B) × (1000/w_A) → 0.2 = (5/M_B) × (1000/50) = 100/M_B

M_B = 100/0.2 = 500 g/mol

✓ Molar mass = 500 g/mol

Trap QuestionNaCl and glucose solutions of same molality have same ΔT_f — True or False?▾

Equal molality NaCl and glucose solutions. A student claims both have the same ΔT_f since ΔT_f = K_f × m.

ΔT_f = i × K_f × m. The van't Hoff factor i matters for electrolytes.

Glucose (non-electrolyte): i = 1. ΔT_f = 1 × K_f × m = K_f·m

NaCl (strong electrolyte): NaCl → Na⁺ + Cl⁻. i ≈ 2 (assuming complete dissociation).

ΔT_f(NaCl) = 2 × K_f × m = 2K_f·m — TWICE as large as glucose.

The trap: forgetting the van't Hoff factor i for electrolytes.

✓ False — NaCl has i ≈ 2 (two ions), so ΔT_f is approximately double that of glucose

Mistake DNA

4 solution errors from EAPCET distractor analysis.

🧊

Freezing Point Elevated (Not Depressed) on Adding Solute

Adding solute LOWERS the freezing point (ΔT_f is a depression). Students confuse with boiling point elevation.

❌ Wrong

Solute added to water: freezing point = 0 + ΔT_f ✗ (freezing point rises!?)

✓ Correct

Freezing point = 0 − ΔT_f ✓ (depressed below 0°C) ✓ Boiling point: elevated ✓ (increased above 100°C)

Adding solute: (1) raises boiling point — harder to vaporise. (2) lowers freezing point — harder to freeze. Anti-freeze in cars uses freezing point depression to prevent water from freezing.

🔢

Using Molarity Instead of Molality in Colligative Property Formulas

ΔT_b = K_b × m uses MOLALITY (moles per kg solvent). Students use molarity (moles per litre solution).

❌ Wrong

ΔT_f = K_f × Molarity ✗ (molarity depends on temperature; wrong unit)

✓ Correct

ΔT_f = K_f × Molality ✓ molality = mol/(kg solvent) ✓ Temperature independent ✓

All colligative property formulas use molality (m) — moles of solute per kilogram of SOLVENT (not solution). Molality is temperature-independent, making it the correct concentration measure for thermodynamic colligative properties.

⚗️

van't Hoff Factor i Ignored for Electrolytes

For electrolytes, all colligative properties are multiplied by the van't Hoff factor i. Forgetting i gives values that are too small by factor i.

❌ Wrong

NaCl ΔT_b calculation: ΔT_b = K_b × m ✗ (forgot i=2 for NaCl)

✓ Correct

ΔT_b = i × K_b × m ✓ i(NaCl) ≈ 2 ✓ ΔT_b = 2 × 0.52 × m ✓

Van't Hoff factor i = number of particles a formula unit produces in solution. NaCl → 2, K₂SO₄ → 3, glucose → 1. All colligative properties for electrolytes: multiply by i.

🫧

Raoult's Law: Partial Pressure = χ × P°_A of SOLVENT (Not Solute)

Raoult's law gives vapour pressure of the SOLVENT above a solution. P_A = χ_A × P°_A where A is the solvent.

❌ Wrong

Vapour pressure of solution: P = χ_solute × P°_solvent ✗ (wrong mole fraction)

✓ Correct

P_solvent = χ_solvent × P°_solvent ✓ Relative lowering: ΔP/P° = χ_solute ✓

Raoult's law: P_A = χ_A × P°_A, where A = solvent, χ_A = mole fraction of SOLVENT. The relative lowering ΔP/P° = χ_B (mole fraction of solute) — a neat result showing the depression is proportional to solute concentration.

Chapter Intelligence

Solutions is numerically predictable — the same four formulas appear year after year in EAPCET.

EAPCET Weightage (2019–2024)

Boiling point elevation

Freezing point depression

Osmotic pressure (van't Hoff)

Relative VP lowering (Raoult's)

van't Hoff factor (i)

High-Yield PYQ Patterns

ΔT_b = K_b·m calculationFind molar mass from ΔT_fOsmotic pressure π = CRTRelative VP lowering = χ_soluteEffect of i for NaCl vs glucoseIsotonic solutions (equal π)Reverse osmosis condition

Exam Strategy

All four colligative properties use molality, not molarity. Convert: m = (g_solute/M_solute) × (1000/g_solvent).
Electrolytes have i > 1 (dissociation gives more particles). For NaCl i≈2, K₂SO₄ i≈3, MgSO₄ i≈2. Always include i for salts.
Finding molar mass: set up ΔT_f = K_f × (w_B/M_B) × (1000/w_A), plug in and solve for M_B. This is a guaranteed 1–2 marks per EAPCET.
Osmotic pressure is the most sensitive colligative property (can detect small concentrations). π = CRT at 27°C (T=300K).
Solutions connects to Equilibrium (Raoult's law derivation), Electrochemistry (electrolyte solutions), and Kinetics (concentration effects on rate).