Numerical Problems Based on Class 12 Chemistry Chapter 3 Electrochemistry

Numerical Problems Based on Class 12 Chemistry Chapter 3 Electrochemistry

Q.1. Calculate the resistance of 0.01 N solution of an electrolyte whose equivalent conductivity is 420 ohm^-1 cm² equiv^-1 (The cell constant of the cell is 0.88 cm^-1).

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Answer: 209.5 ohm

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Q.2. A conductivity cell when filled with 0.02 M KCl (conductivity = 0.002768 cm^-1) has a resistance of 457.3 2. What will be the equivalent conductivity of 0.05 N CaCl₂ solution if the same cell filled with this solution has a resistance of 202 Ω?

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Answer: 125.4 ohm-1 cm2 equiv-1

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Q.3. The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9 S cm² mol^-1, Calculate the conductivity of this solution.

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Answer: 0.20835 S cm^-1

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Q.4. The resistance of a 0.5 M solution of an electrolyte enclosed between two platinum electrodes 1.5 cm apart and having an area of 2.0 cm² was found to be 30 ohm. Calculate the molar conductivity of the solution.

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Answer: 50 ohm-1 mol-1

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Q.5. The resistance of 0.5 N solution of an electrolyte in a conductivity cell was found to be 25 ohm. Calculate the equivalent conductivity of the solution if the electrodes in the cell are 1.6 cm apart and have an area of 3.2 cm².

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Answer: 40

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Q.6. When a certain conductance cell was filled with 0.20 mol dm^-3 aqueous KCI solution, its conductivity was 2.78 x 10^-3 S cm^-1 and had a resistance 82.5 ohm at 300 K. Calculate the cell constant of the cell.

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Answer: 0.229 cm^-1

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Q.7. The conductivity of a solution containing 1.0 g of anhydrous BaCl₂ in 200 cm³ of the solution has been found to 0.0058 S cm^-1. Calculate the molar conductivity and equivalent conductivity of the solution.

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Answer: Λm = 241.67 S cm² mol^-1
Λ = 120.83 S cm² equiv^-1

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Q.8. Which of the following solutions has larger molar conductivity of an electrolyte?
(i) Solution containing 0.08 mol L^-1 and conductivity equal to 2.0 × 10^-2 ohm^-1cm^-1,
(ii) Solution containing 0.1 mol L of electrolyte and resistivity equal to 50 ohm cm.

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Answer: (i) 250 (ii) 200
Solution (i) has higher molar concentration.

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Q.9. Specific conductivity of N/35 KCl at 298 K is 0.002768 ohm^-1 cm^-1 and it has resistance of 520 ohm. A N/25 solution of a salt kept in the same cell was found to have a resistance of 300 ohm at 298 K. Calculate equivalent conductance of the solution

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Answer: 119.5 S cm² equiv^-1

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Q.10. When a certain conductance cell was filled with 0.1 mol L^-1 KCl, it has a resistance of 85 Ω at 25°C. When the same cell was filled with an aqueous solution of 0.052 mol L^-1 of an electrolyte solution, the resistance was 96 2. Calculate the molar conductivity of the electrolyte at this concentration. (Conductivity of 0.1 mol L^-1 KCl solution is 1.29 x 10^-2 S cm^-1)

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Answer: 219.65 S cm² mol^-1

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Q.11. The resistance of a conductivity cell with 0.1 M KCl solution is found to be 200 Ω at 298 K. When the same cell was filled with 0.02 M NaCl solution, the resistance at the same temperature is found to be 1100 Ω
Calculate:
(i) the cell constant of the cell in m^-1.
(ii) the molar conductivity of 0.02 M NaCl solution in S m² mol^-1.
Given: Conductivity of 0.1 M KCl solution at 298 K= 1.29 S m^-1.

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Answer: (i) 258 m^-1
(ii) 1.175 x 10^-2 S m² mol^-1

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Q.12. The molar conductance of 0.05 M solution of MgCl, is 194.5 Ω cm² mol^-1 at 25°C. A cell with electrodes having 1.50 cm² surface area and 0.50 cm apart is filled with 0.05 M solution of MgCl₂. How much current will flow when the potential difference between the electrodes is 5.0V?

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Answer: 0.146 A

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