Conductivity of 0.02 M, KCl solution is `0.002765 S cm^-1` . If the resistance of the cell containing the solution is 400 ohm, calculate the, cell constant.

The conductivity of N/10 KCl solution at 20^circ C is 0.0212 cdot ohm^-1 ~cm^-1 and the resistance of the cell containing this solution at 20^circ C is 55 ohm. The cell constant is : 4.616 ~cm^-1 , 1.166 ~cm^-1 , 2.173 ~cm^-1 , 3.324 ~cm^-1

Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1M is 100ohm. The conductivity of this solution is 1.29 S m^(-1) . Resistance of the same cell when filled with 0.2M of the same solution is 520ohm. The molar conductivity of 0.02M solution of the electrolyte will be

Conductivity of 0.01 M KCI solution is x S cm^(-1) . When conductivity cell is filled with 0.01 M KCl the conductance observed is y S. When the same cell is filled with 0.01M H_(2)SO_(4) the observed conductance is z S cm^(-1) . Hence conductivity of 0.01 M H_(2)SO_(4) is

The conductivity of 0: 20^prime M solution of K C l at 298 ~K is 0.0248 Scm^-1 . Calculate its molar conductivity.

Conductivity of 0.02 M KCl 25^(@)C is 4xx10^(-3)ohm^(-1)cm^(-1) . The resistance of this solution at 25^(@)C when meaured with a particular cell was 200 ohms. The resistance of 0.01 M CuSO_(4) at 25^(@)C measured with the same cell was 8xx10^(3) ohms. WHat is molar conductivity of he copper sulphate solution?

The resistance of a conductivity cell containing 0.001 M^circ KCl solution at 298 ~K is 1500 Omega . What is the cell constant if conductivity.of 0.001 M KCl solution at 298 K is 0.146 xx 10^-3 S cm^-1

The specific conductivity of 0.1 M solution of a weak acid at 298 K is 0.00159 "ohm^(-1) cm^(-1) . Calculate the degree of dissociation of the acid if its molar conductance at infinite dilution is 350.5 "ohm"^(-1) cm^(2) mol^(-1) .