The resistance and conductivity of 0.02 M KCl solution are 82.4 ohm and 0.002768Scm−1, respectively. When filled with 0.005NK2SO4, the solution had a resistance of 324 ohm. Calculate equivalent conductivity of K2SO4 solution.
A
381.6Scm2mol−1
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B
581.6Scm2mol−1
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C
681.6Scm2mol−1
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D
281.6Scm2mol−1
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Solution
The correct option is D281.6Scm2mol−1 ∧eq=k×1000N=7.04×10−4×10000.0052 =281.6Scm2mol−1
[n factor for K2SO4=2;M=N2]