Equivalent conductivity of $Cr_2(SO_4)_3$ is related to molar conductivity by the expression

At $25^{\circ} C$,the molar conductances at infinite dilution for the strong electrolytes $NaOH$,$NaCl$ and $BaCl_2$ are $248 \times 10^{-4}$,$126 \times 10^{-4}$ and $280 \times 10^{-4} \ S \ m^2 \ mol^{-1}$ respectively. The value of $\lambda_m^{\circ} Ba(OH)_2$ in $S \ m^2 \ mol^{-1}$ is:

Ionic conductance of $Ga^{3+}$ ions and $NO_3^-$ ions at infinite dilution are $120 \, \Omega^{-1} \, cm^2 \, mol^{-1}$ and $50 \, \Omega^{-1} \, cm^2 \, mol^{-1}$ respectively. Equivalent conductance of $Ga(NO_3)_3$ at infinite dilution will be ........... $\Omega^{-1} \, cm^2 \, eq^{-1}$.

What benefit do we get by forming a thin layer of $Pt$ particles on the electrodes of a conductivity cell?

The specific conductivity of $N/10$ $KCl$ solution at $20\,^oC$ is $0.0212\,ohm^{-1}\,cm^{-1}$ and the resistance of the cell containing this solution at $20\,^oC$ is $55\,ohm$. The cell constant is ............. $cm^{-1}$.

The resistance of a decimolar solution of $NaCl$ is $30 \ \Omega$. Calculate the conductivity of the solution if the cell constant is $0.33 \ cm^{-1}$.