Question

The specific conductance of 0.05 N solution of an electrolyte at 298 K is 0.002 S cm^(-1). Calculate the equivalent conductance.

Answer 1

The specific conductance, `k =0.002 " S " cm^(-1)=0.002 " oh "m^(-1) cm^(-1)`
Equivalent concentration (C ) =0.05 N =0.05 g eq. `L^(-1)`
`=(0.05 g" eq".)/(1 L)=(0.05 g" eq".)/(1000 cm^(3))=0.05xx10^(-3) f" eq". cm^(-3)`
`"Equivalent conductance", (Lambda_(E))=(k)/(C)=((0.002 " oh "m^(-1) cm^(-1)))/((0.05xx10^(-3)g " eq ". cm^(-3)))`
`=40 " oh "m^(-1) cm^(2)(g " eq".)^(-1)`.