90 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
(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 . (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
Moles of NaCl = 0.25 × 0.4 = 0.1 mol . Mass = 0.1 × 58.5 = 5.85 g .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
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
Moles of solute = (6/120) = 0.05 mol . Volume = 300 mL = 0.3 L. Molarity = (0.05/0.3) ≈ 0.1667 M .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law
Moles = 0.4 × 0.25 = 0.1 mol . Mass of KNO₃ = 0.1 × 101 = 10.1 g . Total mass = 250 × 1.1 = 275 g . Mass % = (10.1/275) × 100 ≈ 3.673% .
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
Moles of NaOH = (4/40) = 0.1 mol . Volume = 250 mL = 0.25 L. Molarity = (0.1/0.25) = 0.4 M .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Relative Lowering and Elevation of Boiling Point
Moles of methanol = (16/32) = 0.5 mol . Moles of water = (72/18) = 4 mol . Total moles = 0.5 + 4 = 4.5 . Mole fraction = (0.5/4.5) ≈ 0.111 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Relative Lowering and Elevation of Boiling Point
Initial moles of ethanol = (46/46) = 1 . Moles of water = (54/18) = 3 . Let additional moles of ethanol = x . New mole fraction = (1 + x/1 + x + 3) = 0.4 . 1 + x = 0.4 (4 + x) , 1 + x = 1.6 + 0.4x , 0.6x = 0.6 , x = 1 . Mass added = 1 × 46 = 46 g .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Relative Lowering and Elevation of Boiling Point
Moles of KCl = 0.4 × 0.25 = 0.1 mol . Mass = 0.1 × 74.5 = 7.45 g .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Relative Lowering and Elevation of Boiling Point
