The resistance of 0.01 N solution of an electrolyte is 210 Ω at 298 K with a cell constant of 0.88 cm−1. Calculate the conductivity and equivalent conductivity of the solution.
A
4.19×10−3ohm−1cm−1,419.05Scm2eq−1
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B
5.19×10−3ohm−1cm−1,519.05Scm2eq−1
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C
6.19×10−3ohm−1cm−1,619.05Scm2eq−1
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D
7.19×10−3ohm−1cm−1,719.05Scm2eq−1
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Solution
The correct option is A4.19×10−3ohm−1cm−1,419.05Scm2eq−1 First check what is given in the question. R=210ohmand1a=0.88cm−1Now.K=1R×1a=1210×0.88=4.19×10−3ohm−1cm−1orScm−1∧eq=k×1000N=4.19×10−3×10000.01=419.05Scm2eq−1