#weak electrolyte

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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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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Corrosion and Applications of Electrochemistry

A weak electrolyte has Lambdam° = 400 S cm² mol⁻¹ and Lambdam = 40 S cm² mol⁻¹ at 0.1 M. What is the dissociation consta

α = (Lambdam/Lambdam°) = (40/400) = 0.1 . Ka = (α² c/1 - α) = (0.1² × 0.1/1 - 0.1) = (0.01 × 0.1/0.9) = 1.111 × 10⁻³ .

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

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Corrosion and Applications of Electrochemistry

A weak electrolyte has Lambdam° = 400 S cm² mol⁻¹ and Lambdam = 40 S cm² mol⁻¹ at 0.1 M. What is the dissociation consta

α = (Lambdam/Lambdam°) = (40/400) = 0.1 . Ka = (α² c/1 - α) = (0.1² × 0.1/1 - 0.1) = (0.01 × 0.1/0.9) = 1.111 × 10⁻³ .

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 conductivity of a 0.002 M weak electrolyte solution is 8 × 10⁻⁵ S cm⁻¹. What is its molar conductivity?

Lambdam = (kappa × 1000/c) = (8 × 10⁻⁵ × 1000/0.002) = 40 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 conductivity of a 0.002 M weak electrolyte solution is 8 × 10⁻⁵ S cm⁻¹. What is its molar conductivity?

Lambdam = (kappa × 1000/c) = (8 × 10⁻⁵ × 1000/0.002) = 40 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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