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

The specific conductance of 0.05 N solution of an electrolyte at 298 K

1.	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» 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)`.

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

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