$A$ solution of concentration $C \ g \ equiv/L$ has a specific resistance $R$. The equivalent conductance of the solution is

The cell constant of a cell is $0.5 \ cm^{-1}$ and the resistance $R = 50 \ \Omega$. If the normality is $1 \ N$,then the equivalent conductivity of the electrochemical cell is ...... $\Omega^{-1} \ cm^2 \ (g \ eq)^{-1}$.

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:

The molar conductivities $(\lambda_{m}^0)$ at infinite dilution of $KBr$,$HBr$,and $KNH_2$ are $120.5$,$420.6$,and $90.48 \ S \ cm^2 \ mol^{-1}$ respectively. Find the value of $\lambda_{m}^0$ for $NH_3$.

Resistance of a conductivity cell filled with $0.1 \, mol \, L^{-1}$ $KCl$ solution is $100 \, \Omega$. If the resistance of the same cell when filled with $0.02 \, mol \, L^{-1}$ $KCl$ solution is $520 \, \Omega$,calculate the conductivity and molar conductivity of $0.02 \, mol \, L^{-1} \, KCl$ solution. The conductivity of $0.1 \, mol \, L^{-1} \, KCl$ solution is $1.29 \, S / m$.

When a solution of an electrolyte is heated,the conductance of the solution: