The resistance of a $0.1 \ N$ solution of acetic acid is $250 \ \Omega$ and the cell constant is $1.15 \ cm^{-1}$. The equivalent conductance of $0.1 \ N$ acetic acid in $\Omega^{-1} \ cm^2 \ eq^{-1}$ is:

$\Lambda_{m}^0$ for $NaCl$,$HCl$ and $NaAc$ are $126.4 \ S \ cm^2 \ mol^{-1}$,$425.9 \ S \ cm^2 \ mol^{-1}$ and $91.0 \ S \ cm^2 \ mol^{-1}$ respectively. Calculate $\Lambda^0$ for $HAc$.

The resistance of a $0.5 \, M$ solution of an electrolyte in a cell was found to be $50 \, \Omega$. If the electrodes in the cell are $2.2 \, cm$ apart and have an area of $4.4 \, cm^2$,then the molar conductivity (in $S \, m^2 \, mol^{-1}$) of the solution is:

The limiting molar conductivities of the given electrolytes at $298 \ K$ follow the order: $\lambda^{\circ}_{(K^{+})} = 73.5, \lambda^{\circ}_{(Cl^{-})} = 76.3, \lambda^{\circ}_{(Ca^{2+})} = 119.0, \lambda^{\circ}_{(SO_{4}^{2-})} = 160.0 \ S \ cm^{2} \ mol^{-1}$.

If the molar conductivity of a solution is $1.26 \times 10^{2} \ \Omega^{-1} \ cm^{2} \ mol^{-1}$ and its molarity is $0.01 \ M$,what will be its specific conductivity?

The correct order of equivalent conductances at infinite dilution in water at room temperature for $H^{+}$, $K^{+}$, $CH_{3}COO^{-}$ and $HO^{-}$ ions is