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

The molar conductivity of $KCl$ solutions at different concentrations at $298 \, K$ are given below:

$c^{1/2} / (mol \, L^{-1})^{1/2}$	$\Lambda_m / S \, cm^2 \, mol^{-1}$
$0.000198$	$148.61$
$0.000309$	$148.29$
$0.000521$	$147.81$
$0.000989$	$147.09$

Show that a plot between $\Lambda_m$ and $c^{1/2}$ is a straight line. Determine the values of $\Lambda_m^o$ and $A$ for $KCl$.

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 molar conductivities (in $S \ cm^2 \ mol^{-1}$) of $NaCl$,$KCl$,and $NaOH$ at infinite dilution are $126$,$150$,and $250$,respectively,the molar conductivity of $KOH$ (in $S \ cm^2 \ mol^{-1}$) is

Why does the conductivity of a solution decrease with dilution?

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 $NaBr$?