Solutions Class 12 Notes — CBSE Chemistry Chapter 2

Chapter 2 of Class 12 Chemistry deals with Solutions — one of the most important chapters for both Boards and competitive exams. This chapter covers how substances dissolve, how to express concentration, and the fascinating colligative properties that depend only on the number of solute particles, not their identity. Expect 5-7 marks from this chapter, mainly from numericals on colligative properties.

Key Concepts

Types of Solutions

SoluteSolventTypeExample
GasGasGaseousAir (O₂ in N₂)
GasLiquidLiquidSoda water (CO₂ in water)
LiquidLiquidLiquidEthanol in water
SolidLiquidLiquidSugar in water
SolidSolidSolidAlloys (brass = Cu + Zn)

Expressing Concentration

Molarity (M) = moles of solute / volume of solution (L)
Molality (m) = moles of solute / mass of solvent (kg)
Mole Fraction (x) = moles of component / total moles
Mass % = (mass of solute / mass of solution) × 100
ppm = (mass of solute / mass of solution) × 10⁶
Why Molality is preferred: Molality doesn’t change with temperature (it depends on mass, not volume), while Molarity does. This is why colligative property formulas use molality!

Solubility

Henry’s Law (Gas in Liquid)

p = K_H × x
Where: p = partial pressure of gas, K_H = Henry’s law constant, x = mole fraction of gas in solution
Higher K_H → lower solubility

Applications:

  • Soda/cold drinks: CO₂ dissolved under high pressure (low solubility at atmospheric pressure → fizz)
  • Scuba diving: High pressure → more N₂ dissolves in blood → “bends” on rapid ascent
  • Oxygen dissolved at higher altitude decreases → altitude sickness

Factors Affecting Solubility

  • Gas in Liquid: Solubility increases with pressure, decreases with temperature
  • Solid in Liquid: Solubility generally increases with temperature (exceptions: CeSO₄, Li₂CO₃)

Raoult’s Law

For volatile solute: p₁ = x₁ · p₁⁰ and p₂ = x₂ · p₂⁰
Total pressure: P_total = p₁ + p₂ = x₁p₁⁰ + x₂p₂⁰

For non-volatile solute: p = x_solvent · p⁰_solvent
Or: (p⁰ – p)/p⁰ = x_solute (relative lowering of vapour pressure)

Ideal vs Non-ideal Solutions

PropertyIdeal SolutionPositive DeviationNegative Deviation
Obeys Raoult’s Law?YesP_observed > P_RaoultP_observed < P_Raoult
ΔH_mix0Positive (endothermic)Negative (exothermic)
ΔV_mix0PositiveNegative
A-B interactions vs A-A, B-BEqualWeakerStronger
ExamplesBenzene + Toluene, n-hexane + n-heptaneEthanol + Water, Acetone + CS₂CHCl₃ + Acetone, HNO₃ + Water
Azeotropes: Solutions that boil at constant temperature and cannot be separated by distillation. Minimum boiling azeotrope → positive deviation (e.g., ethanol-water at 95.4%). Maximum boiling azeotrope → negative deviation (e.g., HNO₃-water at 68%).

Colligative Properties

Properties that depend only on the number of solute particles, not their nature:

1. Relative Lowering of Vapour Pressure (RLVP)

(p⁰ – p)/p⁰ = x_solute = n₂/(n₁ + n₂) ≈ n₂/n₁ (for dilute solutions)
Where n₂ = moles of solute, n₁ = moles of solvent

2. Elevation of Boiling Point (ΔT_b)

ΔT_b = K_b × m
K_b = molal elevation constant (ebullioscopic constant)
m = molality
ΔT_b = (K_b × w₂ × 1000) / (M₂ × w₁) (for finding molar mass M₂)

3. Depression of Freezing Point (ΔT_f)

ΔT_f = K_f × m
K_f = molal depression constant (cryoscopic constant)
ΔT_f = (K_f × w₂ × 1000) / (M₂ × w₁)
Practical Application: This is why we add salt to icy roads — salt lowers the freezing point of water! Also why antifreeze (ethylene glycol) is added to car radiators in cold regions.

4. Osmotic Pressure (π)

π = CRT = (n/V)RT
C = molarity, R = 0.0821 L·atm/mol·K, T = temperature in K
For molar mass: M₂ = (w₂RT) / (πV)

Osmosis: Flow of solvent from lower concentration to higher concentration through a semipermeable membrane.

  • Isotonic solutions: Same osmotic pressure (e.g., saline drip = 0.9% NaCl)
  • Hypertonic: Higher osmotic pressure → cells shrink (crenation)
  • Hypotonic: Lower osmotic pressure → cells swell/burst (haemolysis)
  • Reverse osmosis: Apply pressure > π to force solvent from concentrated to dilute side (used in water purification)

van’t Hoff Factor (i)

i = observed colligative property / calculated colligative property
i = total particles after dissociation or association / particles before

For electrolytes that dissociate: i > 1 (e.g., NaCl → i ≈ 2, CaCl₂ → i ≈ 3)
For solutes that associate: i < 1 (e.g., acetic acid in benzene → i ≈ 0.5)

Modified formulas: ΔT_b = i·K_b·m, ΔT_f = i·K_f·m, π = iCRT

Important Definitions

TermDefinition
SolutionHomogeneous mixture of two or more substances
MolalityMoles of solute per kilogram of solvent
Mole FractionRatio of moles of a component to total moles
Henry’s LawPartial pressure of a gas ∝ its mole fraction in solution
Raoult’s LawVapour pressure of a component = mole fraction × pure vapour pressure
Ideal SolutionSolution that obeys Raoult’s law at all concentrations
AzeotropeConstant-boiling mixture that cannot be separated by distillation
Colligative PropertyProperty depending only on number of solute particles, not their nature
Osmotic PressureMinimum pressure needed to prevent osmosis
van’t Hoff FactorRatio of observed to calculated colligative property

Solved Examples — NCERT Based

Example 1: Molality Calculation

Q: Calculate the molality of a solution containing 20 g of NaOH dissolved in 500 g of water. (M of NaOH = 40)

Solution:

Moles of NaOH = 20/40 = 0.5 mol

Mass of solvent = 500 g = 0.5 kg

Molality = 0.5/0.5 = 1 m (molal)

Example 2: Boiling Point Elevation

Q: 18 g of glucose (M = 180) is dissolved in 500 g of water. Calculate the boiling point of the solution. (K_b for water = 0.52 K·kg/mol)

Solution:

Moles of glucose = 18/180 = 0.1 mol

Molality = 0.1/0.5 = 0.2 m

ΔT_b = K_b × m = 0.52 × 0.2 = 0.104 K

Boiling point = 100 + 0.104 = 100.104°C

Example 3: Osmotic Pressure

Q: 200 mL of an aqueous solution of a protein contains 1.26 g of protein. The osmotic pressure at 300 K is 2.57 × 10⁻³ atm. Calculate the molar mass of the protein.

Solution:

π = CRT → C = π/RT = (2.57 × 10⁻³) / (0.0821 × 300) = 1.044 × 10⁻⁴ mol/L

Moles = C × V = 1.044 × 10⁻⁴ × 0.2 = 2.088 × 10⁻⁵ mol

M = mass/moles = 1.26 / (2.088 × 10⁻⁵) = 60,345 g/mol ≈ 6.03 × 10⁴ g/mol

Example 4: van’t Hoff Factor

Q: 0.1 m NaCl solution has a freezing point depression of 0.348°C. Calculate the van’t Hoff factor. (K_f = 1.86 K·kg/mol)

Solution:

Expected ΔT_f = K_f × m = 1.86 × 0.1 = 0.186°C

i = Observed ΔT_f / Expected ΔT_f = 0.348 / 0.186 = 1.87

(Close to 2, as expected for NaCl → Na⁺ + Cl⁻, though not fully dissociated)

Important Questions for Board Exams

1 Mark Questions

  1. Define molality.
  2. State Henry’s law.
  3. What are isotonic solutions?
  4. Why is osmotic pressure preferred for molar mass of polymers?
  5. What is the van’t Hoff factor for Na₂SO₄ if completely dissociated?

2 Mark Questions

  1. Distinguish between ideal and non-ideal solutions.
  2. Why do gases become less soluble at higher temperatures?
  3. Explain reverse osmosis and its application.
  4. Calculate mole fraction of solute and solvent if 10 g of NaCl is dissolved in 90 g of water.

3 Mark Questions

  1. State Raoult’s law. Explain positive and negative deviations with examples.
  2. Derive the relationship between relative lowering of vapour pressure and mole fraction of solute.
  3. Calculate the freezing point of a solution containing 60 g of glucose in 250 g of water.
  4. What are azeotropes? Explain minimum and maximum boiling azeotropes.

5 Mark Questions

  1. Describe all four colligative properties with formulas and examples. How is van’t Hoff factor used to modify them?
  2. What is osmosis? Explain osmotic pressure, its measurement, and applications. Define isotonic, hypertonic and hypotonic solutions.

Quick Revision Points

  • Molality is temperature-independent; molarity is not
  • Henry’s law: p = K_H × x (higher K_H → less soluble)
  • Raoult’s law: p = x × p⁰ (ideal solutions obey this)
  • Positive deviation: weaker A-B forces, ΔH > 0 | Negative: stronger A-B, ΔH < 0
  • 4 colligative properties: RLVP, ΔT_b, ΔT_f, π — all depend on number of particles only
  • ΔT_b = K_b·m | ΔT_f = K_f·m | π = CRT
  • For electrolytes: multiply by van’t Hoff factor (i)
  • NaCl: i ≈ 2 | CaCl₂: i ≈ 3 | Glucose: i = 1
  • Osmotic pressure is best for molar mass of macromolecules (measurable even at low conc.)
  • Reverse osmosis: pressure > π → pure water through membrane

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top