#degree of dissociation

Variation of Conductivity with Concentration and Measurement

The molar conductivity of a strong electrolyte at infinite dilution is 240 S cm² mol⁻¹. If its molar conductivity at 0.0

For a strong electrolyte, α = (Lambdam/Lambdam°) , but here it’s less than 1 due to ion interaction. α = (228/240) = 0.95 .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement

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Variation of Conductivity with Concentration and Measurement

What is the degree of dissociation of a weak electrolyte with Lambdam = 46 S cm² mol⁻¹ and Lambdam° = 404 S cm² mol⁻¹ ?

α = (Lambdam/Lambdam°) = (46/404) = 0.114 .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement

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Electrochemical Cells - Galvanic Cells and Electrode Potential

A weak electrolyte has Lambdam° = 390 S cm² mol⁻¹ and a degree of dissociation of 0.02 at 0.1 M. What is the conductivit

Lambdam = α Lambdam° = 0.02 × 390 = 7.8 S cm² mol⁻¹ . kappa = (Lambdam × c/1000) = (7.8 × 0.1/1000) = 0.00078 S cm⁻¹ .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential

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Corrosion and Applications of Electrochemistry

The degree of dissociation of a weak electrolyte is 0.05, and its molar conductivity is 19 S cm² mol⁻¹. What is its Lamb

α = (Lambdam/Lambdam°) , 0.05 = (19/Lambdam°) , Lambdam° = 380 S cm² mol⁻¹ .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry

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Conductance - Conductivity Molar Conductivity and Kohlrausch Law

The degree of dissociation of a weak acid is 0.02, and its molar conductivity at infinite dilution is 400 S cm² mol⁻¹. W

α = (Lambdam/Lambdam°) , 0.02 = (Lambdam/400) , Lambdam = 8 S cm² mol⁻¹ .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Conductance - Conductivity Molar Conductivity and Kohlrausch Law

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Conductance - Conductivity Molar Conductivity and Kohlrausch Law

The degree of dissociation of a weak acid is 0.02, and its molar conductivity at infinite dilution is 400 S cm² mol⁻¹. W

α = (Lambdam/Lambdam°) , 0.02 = (Lambdam/400) , Lambdam = 8 S cm² mol⁻¹ .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Conductance - Conductivity Molar Conductivity and Kohlrausch Law

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Conductance - Conductivity Molar Conductivity and Kohlrausch Law

The degree of dissociation of a weak electrolyte is 0.25, and its molar conductivity is 95 S cm² mol⁻¹. What is its Lamb

α = (Lambdam/Lambdam°) , 0.25 = (95/Lambdam°) , Lambdam° = 380 S cm² mol⁻¹ .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Conductance - Conductivity Molar Conductivity and Kohlrausch Law

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Conductance - Conductivity Molar Conductivity and Kohlrausch Law

The degree of dissociation of a weak acid is 0.1, and its molar conductivity at infinite dilution is 380 S cm² mol⁻¹. Wh

α = (Lambdam/Lambdam°) , 0.1 = (Lambdam/380) , Lambdam = 38 S cm² mol⁻¹ .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Conductance - Conductivity Molar Conductivity and Kohlrausch Law

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Conductance - Conductivity Molar Conductivity and Kohlrausch Law

The degree of dissociation of a weak acid is 0.1, and its molar conductivity at infinite dilution is 380 S cm² mol⁻¹. Wh

α = (Lambdam/Lambdam°) , 0.1 = (Lambdam/380) , Lambdam = 38 S cm² mol⁻¹ .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Conductance - Conductivity Molar Conductivity and Kohlrausch Law

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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 - Relative Lowering and Elevation of Boiling Point

A 0.1 molal solution of a solute in water has a boiling point elevation of 0.156°C. If the solute dissociates into 3 ion

Δ Tb = i · Kb · m . 0.156 = i × 0.52 × 0.1 . i = (0.156/0.52 × 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: Colligative Properties - Relative Lowering and Elevation of Boiling Point

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Colligative Properties - Osmotic Pressure and Reverse Osmosis

A 0.25 M solution of a solute in 400 mL of water has an osmotic pressure of 1.845 atm at 27°C. If the solute dissociates

Pi = i · M · RT . 1.845 = i × 0.25 × 0.0821 × 300 . i = (1.845/0.25 × 24.63) ≈ 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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