60 public questions tagged with this topic.
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
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
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
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
(p⁰ - p/p⁰) = xsolute . (30 - 27/30) = 0.1 . Mole fraction of solvent = 1 - 0.1 = 0.9 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
Liquid phase: 310 = 250 x₁ + 350 (1 - x₁) . 310 = 250 x₁ + 350 - 350 x₁ , 100 x₁ = 40 , x₁ = 0.4 , x₂ = 0.6 . Vapor phase: y₁ = (P₁⁰ · x₁/Ptotal) = (250 × 0.4/310) ≈ 0.3226 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
(p⁰ - p/p⁰) = xsolute . (32 - 30/32) = (2/32) = 0.0625 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
Liquid phase: 680 = 600 x₁ + 800 (1 - x₁) . 680 = 600 x₁ + 800 - 800 x₁ , 200 x₁ = 120 , x₁ = 0.6 , x₂ = 0.4 . Vapor phase: y₁ = (600 × 0.6/680) ≈ 0.5294 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
Liquid phase: 580 = 500 x₁ + 700 (1 - x₁) . 580 = 500 x₁ + 700 - 700 x₁ , 200 x₁ = 120 , x₁ = 0.6 , x₂ = 0.4 . Vapor phase: y₁ = (500 × 0.6/580) ≈ 0.5172 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
