57 public questions tagged with this topic.
Relative lowering = (p⁰ - p/p⁰) . (24 - 22.8/24) = (1.2/24) = 0.05 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
(p⁰ - p/p⁰) = xsolute . (28 - 26.6/28) = (1.4/28) = 0.05 . Moles of water = (360/18) = 20 . xsolute = (nsolute/nsolute + 20) = 0.05 . nsolute = 0.05 (nsolute + 20) , 0.95 nsolute = 1 , nsolute ≈ 1.0526 . Mass = 1.0526 × 50 ≈ 52.63 g .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
Δ Tf = Kf · m . 0.372 = 1.86 · m . m = (0.372/1.86) = 0.2 mol/kg . Cross-check: xsolute = (25 - 23.7/25) = 0.052 , consistent for dilute solution.
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
Δ Tb = 100.208 - 100 = 0.208 K . Δ Tb = Kb · m . 0.208 = 0.52 · m , m = (0.208/0.52) = 0.4 mol/kg . Cross-check: xsolute = (24 - 22.4/24) = 0.0667 , consistent for dilute solution.
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
(p⁰ - p/p⁰) = xsolute . Initial: (50 - 47/50) = 0.06 . Doubling molality doubles xsolute (approximately for dilute solutions), so new xsolute = 0.12 . New p = p⁰ (1 - xsolute) = 50 (1 - 0.12) = 44 mm Hg .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
Pi = i · M · RT . Molarity ≈ molality for dilute solution, so M = 0.05 M . Pi = 2.7 × 0.05 × 0.0821 × 300 ≈ 3.326 atm .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Types of Solutions and Expressing Concentration
Moles = (10/40) = 0.25 . Molality = (0.25/0.4) = 0.625 mol/kg . Δ Tf = i · Kf · m . 0.93 = i × 1.86 × 0.625 , i = (0.93/1.86 × 0.625) ≈ 0.8 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Types of Solutions and Expressing Concentration
(p⁰ - p/p⁰) = xsolute . (25 - 24/25) = 0.04 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Types of Solutions and Expressing Concentration
Pi = i · M · RT . Molarity ≈ molality for dilute solution, so M = 0.1 M . 2.46 = i × 0.1 × 0.0821 × 300 . i = (2.46/0.1 × 0.0821 × 300) ≈ 1 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Types of Solutions and Expressing Concentration
Moles = (8/80) = 0.1 . Molality = (0.1/0.5) = 0.2 mol/kg . Δ Tf = i · Kf · m . 0.372 = i × 1.86 × 0.2 , i = (0.372/1.86 × 0.2) = 1 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Types of Solutions and Expressing Concentration
