Explores different solution types (solid, liquid, gas) and methods to express concentration such as molarity, molality, and mass percent.
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
Mole fraction of solvent = 1 - 0.25 = 0.75 . Vapor pressure = 40 × 0.75 = 30 mm Hg .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Types of Solutions and Expressing Concentration
Total mass = 5 g + 45 g = 50 g. Mass % = (5/50) × 100 = 10% .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Types of Solutions and Expressing Concentration
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
(p⁰ - p/p⁰) = xsolute . (25 - 22/25) = (3/25) = 0.12 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Types of Solutions and Expressing Concentration
Moles of CaCl₂ = 0.2 × 0.6 = 0.12 mol . Mass = 0.12 × 111 = 13.32 g .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Types of Solutions and Expressing Concentration
(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
