The molar conductivity of a $0.02 \ M$ solution of an electrolyte is $124 \times 10^{-4} \ S \ m^2 \ mol^{-1}$. What is the resistance of the same solution (in ohms),kept in a cell with a cell constant of $129 \ m^{-1}$?

The conductivity of a centimolar solution of $KCl$ at $25^{\circ} C$ is $0.0210 \, \Omega^{-1} cm^{-1}$ and the resistance of the cell containing the solution at $25^{\circ} C$ is $60 \, \Omega$. The value of the cell constant is $......... \, cm^{-1}$.

Equivalent conductance of an electrolyte containing $NaF$ at infinite dilution is $90.1 \, \Omega^{-1} \, cm^2$. If $NaF$ is replaced by $KF$,what is the value of equivalent conductance? ........... $\Omega^{-1} \, cm^{2}$

Conductivity of $0.001 \ M$ aqueous solution of $Na_2SO_4$ is found to be $2.6 \times 10^{-3} \ S \ cm^{-1}$ at $25 \ ^oC$. If limiting molar conductance of $Na^+$ is $50 \ S \ cm^2 \ mol^{-1}$,then limiting molar conductance of $SO_4^{2-}$ will be .............. $S \ cm^2 \ mol^{-1}$ (neglect conductivity of water).

The molar conductivities of $KCl$,$NaCl$ and $KNO_3$ are $100$,$120$ and $90 \ S \ cm^2 \ mol^{-1}$ respectively. The molar conductivity of $NaNO_3$ would be .......... $S \ cm^2 \ mol^{-1}$.

$0.5 \ N$ solution of a salt placed between two platinum electrodes $2.0 \ cm$ apart and of area of cross section $4.0 \ cm^2$ has a resistance of $25 \ \Omega$. Calculate the equivalent conductivity of solution ................. $\Omega^{-1} \ cm^2 \ eq^{-1}$