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Mole fraction of water = (0.9/0.9 + 0.1) = 0.9 . Vapor pressure = 25 × 0.9 = 22.5 mm Hg .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
Moles of ethylene glycol = (14/62) ≈ 0.2258 mol . Mass of solvent = 350 g = 0.35 kg. Molality = (0.2258/0.35) ≈ 0.645 mol/kg .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
Liquid phase: 240 = 200 x₁ + 300 (1 - x₁) . 240 = 200 x₁ + 300 - 300 x₁ , 100 x₁ = 60 , x₁ = 0.6 , x₂ = 0.4 . Vapor phase: y₂ = (300 × 0.4/240) = 0.5 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
Relative lowering = (p⁰ - p/p⁰) = xsolute . (20 - 19.8/20) = (0.2/20) = 0.01 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
(p⁰ - p/p⁰) = xsolute . (25 - 23/25) = 0.08 . Moles of water = (180/18) = 10 . xsolute = (nsolute/nsolute + 10) = 0.08 . nsolute = 0.08 (nsolute + 10) , nsolute - 0.08 nsolute = 0.8 , 0.92 nsolute = 0.8 , nsolute ≈ 0.8696 . Mass = 0.8696 × 60 ≈ 52.18 g .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
Mole fraction of water = 1 - 0.1 = 0.9 . Vapor pressure = 50 × 0.9 = 45 mm Hg .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
p = KH · x . x = (8/400) = 0.02 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
Moles of NaOH = (4/40) = 0.1 mol . Molarity = (Moles/Volume in L) , so 0.5 = (0.1/V) . Volume = (0.1/0.5) = 0.2 L = 200 mL . Since total volume is 200 mL, water volume = 200 mL (assuming solute volume is negligible).
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
Raoult’s law: P = P₁⁰ · x₁ + P₂⁰ · (1 - x₁) . P = 400 × 0.6 + 600 × 0.4 = 240 + 240 = 480 mm Hg . Actual = 520 mm Hg ≠ 480 mm Hg, so it does not obey Raoult’s law.
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
Mole fraction of methanol = 1 - 0.3 = 0.7 . Vapor pressure = 90 × 0.7 = 63 mm Hg .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
Mass of solute = (12/100) × 250 = 30 g .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
Δ Tb = Kb · m . 100.39 - 100 = 0.39 . 0.39 = Kb × 0.75 . Kb = (0.39/0.75) = 0.52 K kg mol⁻¹ .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
