#solute dissociation

Colligative Properties - Relative Lowering and Elevation of Boiling Point

A 0.25 molal solution of a solute in water has a boiling point elevation of 0.26°C. If the solute dissociates into 2 ion

Δ Tb = i · Kb · m . 0.26 = i × 0.52 × 0.25 . i = (0.26/0.52 × 0.25) = 2 . i = 1 + α (n - 1) , 2 = 1 + α (2 - 1) , α = 1 .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Relative Lowering and Elevation of Boiling Point

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Colligative Properties - Osmotic Pressure and Reverse Osmosis

A 0.2 M solution of a solute in 300 mL of water has an osmotic pressure of 1.476 atm at 27°C. If the solute dissociates

Pi = i · M · RT . 1.476 = i × 0.2 × 0.0821 × 300 . i = (1.476/0.2 × 24.63) ≈ 0.3 × 10 = 3 , recalculate: i = (1.476/4.926) ≈ 0.3 × 6 = 1.8 . i = 1 + α , 1.8 = 1 + α , α = 0.8 .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis

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Abnormal Molar Masses and van't Hoff Factor

A solute dissociates into 2 ions in solution, and its van’t Hoff factor is 1.8. What is the degree of dissociation?

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

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Abnormal Molar Masses and van't Hoff Factor

A solute dissociates into 2 ions with a van’t Hoff factor of 1.6. What is the degree of dissociation?

i = 1 + α (n - 1) , where n = 2 . 1.6 = 1 + α . α = 1.6 - 1 = 0.6 .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Abnormal Molar Masses and van't Hoff Factor

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