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Initial mass % = (20/20 + 80) × 100 = 20% . New mass % = 40% = (20/20 + w) × 100 . 0.4 (20 + w) = 20 , 20 + w = 50 , w = 30 g .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law
Molality = (moles/mass of solvent in kg) . 0.015 = (n/m) , at p = 3 bar , x = (3/200) = 0.015 (molality ≈ x for dilute solutions). For 0.03 mole: 0.015 = (0.03/m) , m = (0.03/0.015) = 2 kg = 2000 g .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law
Total mass = 25 + 225 = 250 g . Mass % = (25/250) × 100 = 10% .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law
p = KH · x . x = (5/250) = 0.02 . Molality ≈ x for dilute solutions, so m = 0.02 mol/kg . Moles = 0.02 × 1 = 0.02 mol .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Relative Lowering and Elevation of Boiling Point
Total mass = 15 + 85 = 100 g . Mass % = (15/100) × 100 = 15% .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Abnormal Molar Masses and van't Hoff Factor
Total mass = 20 g + 80 g = 100 g. Mass % = (20/100) × 100 = 20% , which matches the given value.
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Abnormal Molar Masses and van't Hoff Factor
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