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NCERT Chemistry Class 12 - Chapter 2: Solutions - Notes

CBSEClass 12Chemistryविलयन

Learning Objectives

  • Understand different types of solutions and concentration terms
  • Study Raoult's law and colligative properties
  • Learn about ideal and non-ideal solutions
  • Calculate molecular mass from colligative properties
  • Understand van't Hoff factor and abnormal molar masses

Key Concepts

Types of Solutions

Solutions can be solid-solid, solid-liquid, liquid-liquid, gas-liquid, etc. Most commonly studied: solid in liquid and liquid-liquid.

Concentration Terms

Molarity (M): moles of solute / L of solution. Temperature dependent (volume changes).

Molality (m): moles of solute / kg of solvent. Temperature independent.

Mole fraction (x): xₐ = nₐ/(nₐ + n_b). Mass %: (mass solute/mass solution) × 100.

ppm: parts per million = (mass solute/mass solution) × 10⁶.

Solubility

Henry's Law: p = KH × x (partial pressure of gas ∝ mole fraction in solution). Applications: deep sea diving (bends), carbonated drinks.

Solubility of gases decreases with temperature increase.

Raoult's Law

For volatile solutes: p_A = x_A × p°_A and p_B = x_B × p°_B. Total pressure: P = p_A + p_B.

For non-volatile solute: p = x_solvent × p°_solvent. Relative lowering of vapour pressure: (p° - p)/p° = x_solute.

Ideal and Non-Ideal Solutions

Ideal: Obey Raoult's law. ΔH_mix = 0, ΔV_mix = 0. Example: benzene + toluene, n-hexane + n-heptane.

Positive deviation: P_total > Raoult's law. Weaker A-B interactions. ΔH > 0, ΔV > 0. Example: ethanol + water, acetone + CS₂. Form minimum boiling azeotrope.

Negative deviation: P_total < Raoult's law. Stronger A-B interactions. ΔH < 0, ΔV < 0. Example: chloroform + acetone, HCl + water. Form maximum boiling azeotrope.

Azeotropes: Constant boiling mixtures that cannot be separated by distillation.

Colligative Properties

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

1. Relative lowering of vapour pressure: (p° - p)/p° = x₂ = n₂/(n₁ + n₂)

2. Elevation of boiling point: ΔT_b = K_b × m (K_b = molal elevation constant).

3. Depression of freezing point: ΔT_f = K_f × m (K_f = molal depression constant/cryoscopic constant).

4. Osmotic pressure: π = CRT = (n₂/V)RT. Used for high molecular mass solutes (polymers, proteins).

K_b = RT²_b M₁/(1000 × ΔH_vap); K_f = RT²_f M₁/(1000 × ΔH_fus).

Determination of Molar Mass

M₂ = K_b × w₂ × 1000 / (ΔT_b × w₁) [from boiling point elevation]

M₂ = K_f × w₂ × 1000 / (ΔT_f × w₁) [from freezing point depression]

M₂ = w₂RT / (πV) [from osmotic pressure]

Abnormal Molar Mass and van't Hoff Factor

van't Hoff factor: i = observed colligative property / calculated colligative property = normal molar mass / observed molar mass.

For dissociation: i > 1 (more particles). For association: i < 1 (fewer particles).

i = 1 + (n-1)α (for dissociation into n ions, α = degree of dissociation).

Modified equations: ΔTb = iKbm, ΔTf = iKfm, π = iCRT.

Summary

Solutions are described by various concentration terms. Raoult's law relates vapour pressure to mole fraction. Deviations lead to azeotropes. Four colligative properties depend on solute particle count. Molar mass is determined from colligative property measurements. The van't Hoff factor accounts for dissociation or association of solutes.

Important Terms

  • Raoult's Law: Vapour pressure of component ∝ its mole fraction
  • Colligative Property: Depends on number of solute particles
  • Azeotrope: Constant boiling mixture
  • Osmotic Pressure: Minimum pressure to prevent osmosis
  • Henry's Law: Gas solubility ∝ partial pressure
  • van't Hoff Factor (i): Accounts for association/dissociation

Quick Revision

  • Raoult: p = x × p°; Henry: p = KH × x
  • ΔTb = iKbm; ΔTf = iKfm; π = iCRT
  • +ve deviation: weaker A-B bonds, min boiling azeotrope
  • -ve deviation: stronger A-B bonds, max boiling azeotrope
  • i > 1: dissociation; i < 1: association; i = 1: no change
  • M₂ = Kf w₂ × 1000 / (ΔTf × w₁)
NCERT Chemistry Class 12 - Chapter 2: Solutions - Notes | EduMunch