The resistivity of a `0.8M` solution of an electrolyte is `5times10^(-3) Omega cm`.Its molar conductivity is ______ `10^(4)Omega^(-1)cm^(2)mol^(-1)`.(Narest integer)

The conductivity of 0.02M solution of NaCl is 2.6 xx 10^(-2)" S cm"^(-1) . What is its molar conductivity ?

At 0.04 M concentration, the molar conductivity of solution of an electrolyte is 5000 Omega^(-1)cm^(2) mol^(-1) while at 0.01 M concentration the value is 5100 Omega^(-1)cm^(2)mol^(-1) . Making necessary assumption (Taking it as strong electrolyte) find the molar conductivity at infinite dilution and write percentage dissociation of strong electrolyte at 0.04 M.

At 0.04 M concentration the molar conductivity of a solution of a electrolyte is 5000Omega^(-1)cm^(2)mol^(-1) while at 0.01 M concentration the value is 5100Omega^(-1)cm^(2)mol^(-1) making necessary assumption (taking it as strong electrolyte) find the molar conductivity at infinite dilution and also determine the degree of dissociation of strong electrolyte at 0.04M.

Specific conductivity of 0.01N H_(2)SO_(4) solution is 6 xx 10^(-3) S cm^(-1) . Its molar conductivity is

The molar conductivity of 0.05 M solution of weak acid is 16.6Omega^(-1)cm^(-2)mol^(-1). Its molar conductivity at infinite dilution is 390.7Omega^(-1)cm^(-2)mol^(-1) . The degree of dissociation of weak acid is

The resistance and specific conductance of 0.1M solution of an electrolyte is 40 Omega and 0.014 S cm^(-1) respectively. For the 0.2M solution of same electrolyte if resistance is 210 Omega,then the molar conductivity (in S cm^2 mol^(-1)) will be

Resistance of 0.2 M solution of an electrolue is 50 Omega . The specific conductance of the solution is 1.4 S m^^(-1) . The resistance of 0.5 M solution of the same electrolyte is 280. Omega . The molar conducitivity of 0.5 M solution of the electrolyte is S m^2 "mol"^(-1) is.

If resistivity of 0.8M KCl solution is 2.5xx10^(3)Omega cm calculate lamda_(m) of the solution