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Question

1.0 N solution of a salt surrounding two platinum electrodes of 2.1 cm apart and 4.2 cm2 in area was found to offer a resistance of 50 ohm. Calculate the equivalent conductivity of the solution.

A
Λeq=0.01ohm1cm2eq1
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B
Λeq=10ohm1cm2eq1
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C
Λeq=105ohm1cm2eq1
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D
Λeq=100ohm1cm2eq1
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Solution

The correct option is B Λeq=10ohm1cm2eq1
Given:
Distance between electrodes, l=2.1 cmArea, A=4.2 cm2Resistance, R=50 ohm

Specific conductance, κ=lA×1R

κ=2.14.2×150=0.01ohm1cm1

Equivalent conductance is the conductivity of a solution containing 1 g-equivalent of the electrolyte.

Λeq=κ V
V is volume of solution containing 1 gequiv

Normality of solutionN
N gequiv of solute present in 1 L solution

1 gequiv of solute present in 1N L solution

Λeq=κN (when volume in litres)

Λeq=κ×1000N (when volume in cm3)

Λeq=0.01×10001

Λeq=10ohm1cm2eq1


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