Calculate molar conductivity at infinite dilution for $NaBr$ if molar conductivity at infinite dilution for $NaCl$,$KBr$ and $KCl$ are $126$,$152$ and $150 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ respectively.

Resistance of a conductivity cell (cell constant $129 \; m^{-1}$) filled with $74.5 \; ppm$ solution of $KCl$ is $100 \; \Omega$ (labelled as solution $1$). When the same cell is filled with $KCl$ solution of $149 \; ppm$,the resistance is $50 \; \Omega$ (labelled as solution $2$). The ratio of molar conductivity of solution $1$ and solution $2$ is i.e.,$\frac{\wedge_{1}}{\wedge_{2}} = x \times 10^{-3}$. The value of $x$ is (Nearest integer). Given,molar mass of $KCl$ is $74.5 \; g \; mol^{-1}$.

The pair of electrolytes that possess the same value for the constant $(A)$ in the Debye-Huckel-Onsager equation,$\Lambda_{m} = \Lambda_{m}^{\circ} - A \sqrt{C}$ is

For which electrolyte the molar conductivity at infinite dilution cannot be obtained graphically?

The conductivity of a strong electrolyte is:

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.