The specific conductance of a 0.01 M solution of KCl is 0.0014 `ohm^(-1) cm^(-1) " at " 25^(@)C`. Its equivalent conductance is:

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

The specific conductivity of 0.5N solution is 0.01287 ohm^(-1) cm^(-1) . What would be its equivalent conductance?

The specific conductivity of 0.5N solution is 0.01287 ohm^(-1) cm^(-1) . What would be its equivalent conductance?

At 25^@C , the specific conductance of 0.01 M alkaline earth metal chloride is 0.000158 "ohm"^(-1) cm^(-1) Calculate the equivalent conductance.

At 25^@C , the specific conductance of 0.01 M alkaline earth metal chloride is 0.000158 "ohm"^(-1) cm^(-1) Calculate the equivalent conductance.

The specific conductance of solution of 0.12 N is 0.024 mho^(-1) cm^(-1) . Determine its equivalent conductance.

The specific conductance of a 0.1 N KCl solution at 23^(@)C is 0.012 ohm^(-1) cm^(-1) . The resistance of cell containing the solution at the same tempreature was found to be 55 ohm . The cell constant will be

The specific conductance of 0.01 (N) KCl solution at 25^(@)C, is "0.00140 ohm"^(-1)."cm"^(-1) . The same conductivity cell when filled with ((N)/(100))NaOH solution, the resistance of the solution was 14.8 ohm. If the resistance of the KCl solution is 25 ohm, what will be the equivalent conductivity of NaOH at 25^(@)C ?