10 public questions tagged with this topic.
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
Mole fraction of second = 1 - 0.6 = 0.4 . Total vapor pressure = 250 × 0.6 + 350 × 0.4 = 150 + 140 = 290 mm Hg .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
Pure P₁⁰ = 400 mm Hg . In solution: P₁ = P₁⁰ · x₁ = 400 × 0.5 = 200 mm Hg . Vapor pressure lowering = P₁⁰ - P₁ = 400 - 200 = 200 mm Hg .
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
P = P₁⁰ · x₁ + P₂⁰ · x₂ , where x₁ + x₂ = 1 . 216 = 180 (1 - x₂) + 240 x₂ . 216 = 180 - 180 x₂ + 240 x₂ . 216 - 180 = 60 x₂ , x₂ = (36/60) = 0.6 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Types of Solutions and Expressing Concentration
Mole fraction of second = 1 - 0.7 = 0.3 . Total vapor pressure = 200 × 0.7 + 300 × 0.3 = 140 + 90 = 230 mm Hg .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Types of Solutions and Expressing Concentration
Liquid phase: 380 = 300 x₁ + 500 (1 - x₁) . 380 = 300 x₁ + 500 - 500 x₁ , 200 x₁ = 120 , x₁ = 0.6 , x₂ = 0.4 . Vapor phase: y₂ = (P₂⁰ · x₂/Ptotal) = (500 × 0.4/380) ≈ 0.5263 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Types of Solutions and Expressing Concentration
P = P₁⁰ · x₁ + P₂⁰ · (1 - x₁) . 260 = 200 x₁ + 300 (1 - x₁) . 260 = 200 x₁ + 300 - 300 x₁ . 260 - 300 = -100 x₁ , x₁ = 0.4 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Types of Solutions and Expressing Concentration
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
