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

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

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

(N/A) The Kohlrausch equation is given by: $\Lambda_m = \Lambda_m^o - A \sqrt{c}$.
To verify this,we calculate $c^{1/2}$ for each concentration:

$c / mol \ L^{-1}$	$c^{1/2} / (mol \ L^{-1})^{1/2}$	$\Lambda_m / S \ cm^2 \ mol^{-1}$
$0.000198$	$0.01407$	$148.61$
$0.000309$	$0.01758$	$148.29$
$0.000521$	$0.02283$	$147.81$
$0.000989$	$0.03145$	$147.09$

Plotting $\Lambda_m$ on the y-axis and $c^{1/2}$ on the x-axis yields a straight line.
From the slope of the line,$A = -\frac{\Delta \Lambda_m}{\Delta c^{1/2}} = -\frac{147.09 - 148.61}{0.03145 - 0.01407} \approx -87.46 \ S \ cm^2 \ mol^{-1} (mol \ L^{-1})^{-1/2}$.
Extrapolating the line to $c^{1/2} = 0$ (y-intercept),we get $\Lambda_m^o \approx 150.0 \ S \ cm^2 \ mol^{-1}$.