1.0N solution of a salt surrounding two platinum electrodes 2.1cm apart and 4.2 sq cm in area was found to offer a resistance of 50ohm. Calculate the equivalent conductivity of the solution.
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Solution
Given, l−2.1cm,a=4.2sqcm,R=50ohm Specific conductance, k=1a⋅1R or k=2.14.2×150=0.01ohm−1cm−1 Equivalent conductivity =k×V V= the volume containing 1g−equivalent=1000mL So, Equivalent conductivity =0.01×1000 =10ohm−1cm2eq−1.