Covers solubility principles for solids and gases in liquids, including Henry's Law and factors affecting dissolution.
Δ Tb = Kb · m . 0.104 = Kb × 0.2 . Kb = (0.104/0.2) = 0.52 K kg mol⁻¹ .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law
p = KH · x . x = (5/500) = 0.01 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law
Δ Tb = Kb · m . 100.52 - 100 = 0.52 · m . m = (0.52/0.52) = 1 mol/kg .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law
Henry’s law: p = KH · x . Initial x₁ = (6/150) = 0.04 , final x₂ = (3/150) = 0.02 . Factor = (x₂/x₁) = (0.02/0.04) = 0.5 (decreases by a factor of 2).
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law
Henry’s law: p = KH · x . p = 50 × 0.04 = 2 bar .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law
Total vapor pressure = p₁⁰ · x₁ + p₂⁰ · x₂ . 300 × 0.4 + 400 × 0.6 = 120 + 240 = 360 mm Hg .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law
Henry’s law: p = KH · x . p = 250 × 0.008 = 2 bar .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law
Raoult’s law: p = p⁰ · xbenzene . 95 = 100 · xbenzene . xbenzene = (95/100) = 0.95 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law
According to Henry’s law, solubility of a gas is directly proportional to its partial pressure.
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law
Henry’s law: p = KH · x . p = 120 × 0.025 = 3 bar .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law
Moles of solute = (6/120) = 0.05 mol . Volume = 300 mL = 0.3 L. Molarity = (0.05/0.3) ≈ 0.1667 M .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law
Henry’s law: p = KH · x . x = (4/80) = 0.05 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law
