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The specific conductance of a 0.12 N sol...

The specific conductance of a 0.12 N solution of an electrolyte is `2.4xx10^(-2) S cm^(-1)`. Calculate its equivalent conductance.

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The correct Answer is:
200 `ohm^(-1)cm^(2)eq^(-1)`
`wedge_(eq)=kappaxx(1000)/("Normality")=0.024Omega^(-1)cm^(-1)xx(1000cm^(3)L^(-1))/(0.12g" "eqL^(-1))=200Omega^(-1)cm^(2)eq^(-1)`.
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