The conductivity of sodium chloride at $298 \ K$ has been determined at different concentrations and the results are given below:

Concentration $/ M$	$0.001$	$0.010$	$0.020$	$0.050$	$0.100$
${10^2} \times \kappa / S \, m^{-1}$	$1.237$	$11.85$	$23.15$	$55.53$	$106.74$

Calculate ${\Lambda _m}$ for all concentrations and draw a plot between ${\Lambda _m}$ and $c^{1/2}$. Find the value of $\Lambda _m^o$.

(A) The molar conductivity ${\Lambda _m}$ is calculated using the formula: ${\Lambda _m} = \frac{\kappa \times 1000}{c}$ where $\kappa$ is in $S \, cm^{-1}$ and $c$ is in $mol \, L^{-1}$.
Given $\kappa$ values are in $S \, m^{-1}$,we convert them to $S \, cm^{-1}$ by multiplying by $10^{-2}$.
$1$. For $c = 0.001 \, M$: $\kappa = 1.237 \times 10^{-4} \, S \, cm^{-1} \implies {\Lambda _m} = \frac{1.237 \times 10^{-4} \times 1000}{0.001} = 123.7 \, S \, cm^2 \, mol^{-1}$. $c^{1/2} = 0.0316 \, M^{1/2}$.
$2$. For $c = 0.010 \, M$: $\kappa = 11.85 \times 10^{-4} \, S \, cm^{-1} \implies {\Lambda _m} = \frac{11.85 \times 10^{-4} \times 1000}{0.010} = 118.5 \, S \, cm^2 \, mol^{-1}$. $c^{1/2} = 0.1000 \, M^{1/2}$.
$3$. For $c = 0.020 \, M$: $\kappa = 23.15 \times 10^{-4} \, S \, cm^{-1} \implies {\Lambda _m} = \frac{23.15 \times 10^{-4} \times 1000}{0.020} = 115.8 \, S \, cm^2 \, mol^{-1}$. $c^{1/2} = 0.1414 \, M^{1/2}$.
$4$. For $c = 0.050 \, M$: $\kappa = 55.53 \times 10^{-4} \, S \, cm^{-1} \implies {\Lambda _m} = \frac{55.53 \times 10^{-4} \times 1000}{0.050} = 111.1 \, S \, cm^2 \, mol^{-1}$. $c^{1/2} = 0.2236 \, M^{1/2}$.
$5$. For $c = 0.100 \, M$: $\kappa = 106.74 \times 10^{-4} \, S \, cm^{-1} \implies {\Lambda _m} = \frac{106.74 \times 10^{-4} \times 1000}{0.100} = 106.7 \, S \, cm^2 \, mol^{-1}$. $c^{1/2} = 0.3162 \, M^{1/2}$.
Plotting ${\Lambda _m}$ vs $c^{1/2}$ and extrapolating to $c^{1/2} = 0$,we find the intercept $\Lambda _m^o \approx 124.0 \, S \, cm^2 \, mol^{-1}$.