Given `l//a=0.5 cm^(-1), R= 50 "ohm", N= 1.0.` The equivalent conductance of the electrolytic cell is .

The specific conductance of a0.5 N solution of an electrolyte at 25^@C is 0.00045 Scm^(-1) . The equivalent conductance of this electrolyte at infinite dilution is 300 S cm^2eq^(-1) . The degree of dissociation of the electrolyte is

The specific conductance of a0.5 N solution of an electrolyte at 25^@C is 0.00045 Scm^(-1) . The equivalent conductance of this electrolyte at infinite dilution is 300 S cm^2eq^(-1) . The degree of dissociation of the electrolyte is

The resistivity of 0.5 N solution of an electrolyte in a conductivity cell was found to be 45 ohms. The equivalent conductance of the same solution is ... if the electrodes in the cell are 2.2 cm part and have an area of 3.8 cm^2

The resistance of 0.01 N solution of an electrolyte waw found to be 210 ohm at 298 K, when a conductivity cell with cell constant 0.66 cm^(-1) is used. The equivalent conductance of solution is:

The resistance of 0.01 N solution of an electrolyte was found to be 220 ohm at 298 K using a conductivity cell with a cell constant of 0. 88cm^(-1) . The value of equivalent conductance of solution is

A 0.1N solution of NaCl has a specific conductance of 0.00112 ohm^-1 cm^-1 . Calculate the equivalent conductance of the solution.