35 public questions tagged with this topic.
Lambdam = (kappa × 1000/c) = (0.00104 × 1000/0.008) = 130 S cm² mol⁻¹ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrolytic Cells and Electrolysis and Faraday's Laws
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
Δ Tb = 100.26 - 100 = 0.26 K . Δ Tb = Kb · m . 0.26 = 0.52 · m , m = 0.5 mol/kg . Cross-check: xsolute = (25 - 23.5/25) = 0.06 , but vapor pressure not needed here.
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
Mole fraction of water = 1 - 0.25 = 0.75 . Vapor pressure = 24 × 0.75 = 18 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 . (35 - 33/35) = (2/35) ≈ 0.0571 .
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
