#KCl solution

Nernst Equation and Gibbs Energy and Equilibrium Constant

What is the cell constant if the resistance of a 0.05 M KCl solution is 100 Ω and its conductivity is 0.0065 S cm⁻¹?

kappa = (cell constant/R) , 0.0065 = (cell constant/100) , cell constant = 0.65 cm⁻¹ .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Nernst Equation and Gibbs Energy and Equilibrium Constant

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Nernst Equation and Gibbs Energy and Equilibrium Constant

The resistance of a conductivity cell with 0.001 M KCl solution is 1500 Ω, and its conductivity is 0.146 × 10⁻³ S cm⁻¹.

kappa = (cell constant/R) , 0.146 × 10⁻³ = (cell constant/1500) , cell constant = 0.219 cm⁻¹.

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Nernst Equation and Gibbs Energy and Equilibrium Constant

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Electrochemical Cells - Galvanic Cells and Electrode Potential

The resistance of a conductivity cell with 0.02 M KCl solution is 400 Ω, and its conductivity is 0.0025 S cm⁻¹. What is

kappa = (cell constant/R) , 0.0025 = (cell constant/400) , cell constant = 1.0 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

How many coulombs are required to deposit 6.75 g of potassium from a KCl solution? (Atomic mass of K = 39 g/mol, F = 965

K⁺ + e⁻ → K . 1 mol K (39 g) requires 1F. Moles = (6.75/39) = 0.1731 mol , Charge = 0.1731 × 96500 = 16700 C .

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

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Corrosion and Applications of Electrochemistry

How many coulombs are required to deposit 6.75 g of potassium from a KCl solution? (Atomic mass of K = 39 g/mol, F = 965

K⁺ + e⁻ → K . 1 mol K (39 g) requires 1F. Moles = (6.75/39) = 0.1731 mol , Charge = 0.1731 × 96500 = 16700 C .

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

A conductivity cell with cell constant 0.6 cm⁻¹ shows a resistance of 200 Ω with a 0.03 M KCl solution. What is the mola

kappa = (cell constant/R) = (0.6/200) = 0.003 S cm⁻¹ . Lambdam = (kappa × 1000/c) = (0.003 × 1000/0.03) = 100 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

A conductivity cell with cell constant 0.6 cm⁻¹ shows a resistance of 200 Ω with a 0.03 M KCl solution. What is the mola

kappa = (cell constant/R) = (0.6/200) = 0.003 S cm⁻¹ . Lambdam = (kappa × 1000/c) = (0.003 × 1000/0.03) = 100 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

A conductivity cell with cell constant 0.5 cm⁻¹ shows a resistance of 200 Ω with a 0.02 M KCl solution. What is the mola

kappa = (cell constant/R) = (0.5/200) = 0.0025 S cm⁻¹ . Lambdam = (kappa × 1000/c) = (0.0025 × 1000/0.02) = 125 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

A conductivity cell with cell constant 0.5 cm⁻¹ shows a resistance of 200 Ω with a 0.02 M KCl solution. What is the mola

kappa = (cell constant/R) = (0.5/200) = 0.0025 S cm⁻¹ . Lambdam = (kappa × 1000/c) = (0.0025 × 1000/0.02) = 125 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 molar conductivity of a 0.1 M KCl solution is 129 S cm² mol⁻¹. If the conductivity of the solution is 0.0129 S cm⁻¹,

Lambdam = (kappa × 1000/c) . 129 = (0.0129 × 1000/c) , so c = (12.9/129) = 0.1 M .

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 molar conductivity of a 0.1 M KCl solution is 129 S cm² mol⁻¹. If the conductivity of the solution is 0.0129 S cm⁻¹,

Lambdam = (kappa × 1000/c) . 129 = (0.0129 × 1000/c) , so c = (12.9/129) = 0.1 M .

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

How many grams of KCl (molar mass = 74.5 g/mol) are needed to prepare 250 mL of a 0.4 M solution?

Moles of KCl = 0.4 × 0.25 = 0.1 mol . Mass = 0.1 × 74.5 = 7.45 g .

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

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