The correct order of equivalent conductance of $LiCl$, $NaCl$ and $KCl$ at infinite dilution is

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 $74.5 \; g \; mol^{-1}$.

If the specific conductance of $1 \, M \, H_2SO_4$ solution is $26 \times 10^{-2} \, S \, cm^{-1}$,then the equivalent conductivity would be

Write the formula to calculate conductivity by using a conductivity cell.

The limiting molar conductivities $(\Lambda_0)$ for $NaCl$,$KBr$ and $KCl$ are $126$,$152$ and $150 \ S \ cm^2 \ mol^{-1}$ respectively. What is the $\Lambda_0$ of $NaBr$?

The resistance of $0.1 \ M$ weak acid $HA$ in a conductivity cell is $2 \times 10^3 \ \Omega$. The cell constant of the cell is $0.78 \ cm^{-1}$ and $\lambda_{m}^{\circ}$ of acid $HA$ is $390 \ S \ cm^2 \ mol^{-1}$. The $pH$ of the solution is