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For 2 ions, i = 1 + α . 1.8 = 1 + α . α = 1.8 - 1 = 0.8 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Abnormal Molar Masses and van't Hoff Factor
Pi = i · M · RT . 2.214 = i × 0.3 × 0.0821 × 300 . i = (2.214/0.3 × 24.63) ≈ 0.3 , recalculate: i = (2.214/7.389) ≈ 0.3 × 10 = 3 . i = 1 + α (3 - 1) , 3 = 1 + 2α , α = 1 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Abnormal Molar Masses and van't Hoff Factor
Moles of water = (500 - x/18) , moles of ethanol = (x/46) . Mole fraction = ((x/46)/(x/46) + (500 - x/18)) = 0.2 . Solving: (x/46) = 0.2 left( (x/46) + (500 - x/18) right) . Simplify: x ≈ 138 g (after cross-multiplication and approximation).
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Abnormal Molar Masses and van't Hoff Factor
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
Moles = 0.6 × 0.2 = 0.12 mol . Mass of NaCl = 0.12 × 58.5 = 7.02 g . Total mass = 200 × 1.05 = 210 g . Mass of water = 210 - 7.02 = 202.98 g = 0.20298 kg . Molality = (0.12/0.20298) ≈ 0.591 mol/kg .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Abnormal Molar Masses and van't Hoff Factor
Pi = i · M · RT . 1.476 = i × 0.15 × 0.0821 × 300 . i = (1.476/0.15 × 0.0821 × 300) ≈ 0.4 , which is incorrect; recalculate: i = (1.476/0.15 × 24.63) ≈ 0.4 × 4 = 1.6 . i = 1 + α , 1.6 = 1 + α , α = 0.6 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Abnormal Molar Masses and van't Hoff Factor
Initial moles of methanol = (32/32) = 1 . Moles of water = (90/18) = 5 . Let additional moles of methanol = x . New mole fraction = (1 + x/1 + x + 5) = 0.3 . 1 + x = 0.3 (6 + x) , 1 + x = 1.8 + 0.3x , 0.7x = 0.8 , x ≈ 1.1429 . Mass added = 1.1429 × 32 ≈ 36.57 g .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Abnormal Molar Masses and van't Hoff Factor
Moles of methanol = (23/32) ≈ 0.7188 . Initial moles of water = (72/18) = 4 . New moles of water = (72 + 64/18) = (136/18) ≈ 7.5556 . Total moles = 0.7188 + 7.5556 ≈ 8.2744 . Mole fraction = (0.7188/8.2744) ≈ 0.0869 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Abnormal Molar Masses and van't Hoff Factor
Initial mass % = (30/30 + 120) × 100 = 20% . New mass % = 25% = (30/30 + w) × 100 . 0.25 (30 + w) = 30 , 30 + w = 120 , w = 90 g .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Abnormal Molar Masses and van't Hoff Factor
Δ Tf = i · Kf · m . 0.558 = i × 1.86 × 0.1 . i = (0.558/1.86 × 0.1) = 3 . i = 1 + α (n - 1) , 3 = 1 + α (3 - 1) , 2α = 2 , α = 1 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Abnormal Molar Masses and van't Hoff Factor
i = 1 + α (n - 1) , where n = 3 . 2.2 = 1 + α (3 - 1) . 2.2 = 1 + 2α , α = (1.2/2) = 0.6 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Abnormal Molar Masses and van't Hoff Factor
Moles of solute = (20/40) = 0.5 mol . Moles of water = (80/18) ≈ 4.444 mol . Total moles = 0.5 + 4.444 = 4.944. Mole fraction of solute = (0.5/4.944) ≈ 0.101 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Abnormal Molar Masses and van't Hoff Factor
