45 public questions tagged with this topic.
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
Δ Tb = Kb · m . 100.39 - 100 = 0.39 . 0.39 = Kb × 0.75 . Kb = (0.39/0.75) = 0.52 K kg mol⁻¹ .
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
(p⁰ - p/p⁰) = xsolute . (30 - 28.5/30) = 0.05 = (nsolute/nsolute + nwater) . Molality = (nsolute/wwater) = 0.5 . Assume wwater = 1 kg , then nsolute = 0.5 . nwater = (1000/18) ≈ 55.56 . xsolute = (0.5/0.5 + 55.56) ≈ 0.0089 , adjust wwater = (0.5 × 18/0.05) = 180 g .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
Molality = (moles/mass of solvent in kg) = (0.02/0.8) = 0.025 mol/kg . Cross-check: x = (7/350) = 0.02 , slightly off due to approximation, but molality is direct.
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
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 . 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
Δ Tf = Kf · m . 0.93 = 1.86 · m . m = (0.93/1.86) = 0.5 mol/kg .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Types of Solutions and Expressing Concentration
Δ Tb = Kb · m . 100.52 - 100 = 0.52 · m . m = (0.52/0.52) = 1 mol/kg .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law
Δ Tb = Kb · m . 100.26 - 100 = 0.52 · m . m = (0.26/0.52) = 0.5 mol/kg .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law
Moles of sorbitol = (16/182) ≈ 0.0879 mol . Mass of solvent = 800 g = 0.8 kg. Molality = (0.0879/0.8) ≈ 0.1099 mol/kg .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law
