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The molar conductivity of 0.05 M of solution of an electrolyte is 200 Ω-1 cm2 mol-1. The resistance offered by a conductivity cell with cell constant (1/3 cm-1) would be _______Ω.
Correct answer is '33.33'. Can you explain this answer?
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The molar conductivity of 0.05 M of solution of an electrolyte is 200 ...
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The molar conductivity of 0.05 M of solution of an electrolyte is 200 ...
Calculation of Resistance Offered by a Conductivity Cell:
Given:
- Molar conductivity of the electrolyte solution = 200 -1 cm2 mol-1
- Concentration of the electrolyte solution = 0.05 M
- Cell constant of the conductivity cell = 1/3 cm-1
Formula:
- Molar conductivity (λ) = K / C
- Resistance (R) = 1 / G
- Conductance (G) = K / L
Where,
- K = conductivity of the solution
- C = concentration of the solution
- L = distance between the electrodes of the cell
Calculation:
- Molar conductivity (λ) = 200 -1 cm2 mol-1
- Concentration (C) = 0.05 M
Therefore, the conductivity (K) of the solution can be calculated as:
K = λ x C
= 200 x 0.05
= 10 S cm-1
- Cell constant (L) = 1/3 cm-1
Therefore, the conductance (G) of the solution can be calculated as:
G = K / L
= 10 / (1/3)
= 30 S
- Resistance (R) = 1 / G
= 1 / 30
= 0.0333 Ω
Therefore, the resistance offered by the conductivity cell is 33.33 Ω.
Given:
- Molar conductivity of the electrolyte solution = 200 -1 cm2 mol-1
- Concentration of the electrolyte solution = 0.05 M
- Cell constant of the conductivity cell = 1/3 cm-1
Formula:
- Molar conductivity (λ) = K / C
- Resistance (R) = 1 / G
- Conductance (G) = K / L
Where,
- K = conductivity of the solution
- C = concentration of the solution
- L = distance between the electrodes of the cell
Calculation:
- Molar conductivity (λ) = 200 -1 cm2 mol-1
- Concentration (C) = 0.05 M
Therefore, the conductivity (K) of the solution can be calculated as:
K = λ x C
= 200 x 0.05
= 10 S cm-1
- Cell constant (L) = 1/3 cm-1
Therefore, the conductance (G) of the solution can be calculated as:
G = K / L
= 10 / (1/3)
= 30 S
- Resistance (R) = 1 / G
= 1 / 30
= 0.0333 Ω
Therefore, the resistance offered by the conductivity cell is 33.33 Ω.
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The molar conductivity of 0.05 M of solution of an electrolyte is 200 Ω-1 cm2 mol-1. The resistance offered by a conductivity cell with cell constant (1/3 cm-1) would be _______Ω.Correct answer is '33.33'. Can you explain this answer?
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