9 public questions tagged with this topic.
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
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
(p⁰ - p/p⁰) = xsolute . (30 - 27/30) = 0.1 . Mole fraction of solvent = 1 - 0.1 = 0.9 .
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
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 . 0.13 = 0.52 × (moles/0.4) . Moles = (0.13 × 0.4/0.52) = 0.1 . Molar mass = (10/0.1) = 100 g/mol .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Relative Lowering and Elevation of Boiling Point
Δ Tf = Kf · m . 0.465 = 1.86 × (moles/0.3) . Moles = (0.465 × 0.3/1.86) ≈ 0.075 . Molar mass = (15/0.075) = 200 g/mol .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis
Pi = M · RT . 0.984 = M × 0.0821 × 300 . M = (0.984/0.0821 × 300) ≈ 0.04 M . Cross-check: Moles = 0.04 × 0.25 = 0.01 , but given 0.02 mole, i = 2 , adjust assumption if needed.
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis
