The resistance of a $0.1 \, N$ solution of a salt is $2.5 \times 10^{3} \, \Omega$. If the cell constant is $1.15 \, cm^{-1}$,what will be the equivalent conductance of the solution in $\Omega^{-1} \, cm^{2} \, eq^{-1}$ (in $.6$)?

Resistance of a conductivity cell at $298 \ K$ for $0.0100 \ M \ KCl$ is $161.8 \ \Omega$. Resistance becomes $190 \ \Omega$ when $0.005 \ M \ NaOH$ solution is filled in this cell. Calculate: $(i)$ cell constant,$(ii)$ specific conductivity for $NaOH$ solution,and $(iii)$ molar conductivity.

In infinite dilution,the equivalent conductances of $Ba^{2+}$ and $Cl^{-}$ are $127$ and $76 \ ohm^{-1} cm^2 eqv^{-1}$ respectively. The equivalent conductivity of $BaCl_2$ at infinite dilution is: (in $.5$)

The cell constant of a given cell is $0.47 \, cm^{-1}$. The resistance of a solution placed in this cell is measured to be $31.6 \, \Omega$. The conductivity of the solution (in $S \, cm^{-1}$ where $S$ has usual meaning) is

The molar conductivity for electrolytes $A$ and $B$ are plotted against $C^{1/2}$ as shown below. Electrolytes $A$ and $B$ respectively are:

The specific conductance of a $0.1 \ N \ KCl$ solution at $23 \ ^oC$ is $0.012 \ \Omega^{-1} \ cm^{-1}$. The resistance of a cell containing the solution at the same temperature was found to be $55 \ \Omega$. The cell constant will be .............. $cm^{-1}$.