1.

Specific conductivity of a 0.12 normal solution of an electrolyte is 0.023 ohm^(-1)cm^(-1). Determine its equivalent conductivity.

Answer»


Solution :`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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