A $5mmold{m}^{-3}$ aqueous solution of $KCl$ has a conductance of $0.55mS$ when measured in a cell of cell constant $1.3c{m}^{-1}$. The molar conductivity of this solution is ___________ $mS{m}^{2}mo{l}^{-1}$. (Round off to the nearest integer).

Step 1: Calculate the conductivity of the solution

Conductivity, $K=\frac{Conduc\tan ce}{Cellcons\tan t}=\frac{0.55\times {10}^{-3}}{1.3}$

$\Rightarrow K=71.5\times {10}^{-5}Sc{m}^{-1}$

Step 2: Calculate the molar conductivity of the solution

Molar conductivity, ${\lambda }_m=\frac{K\times 1000}{Molarity}=\frac{71.5\times {10}^{-5}\times 1000}{5\times {10}^{-3}}$

$$\begin{gathered} =143Sc{m}^{2}mo{l}^{-1} \\ =14.3mS{m}^{2}mo{l}^{-1} \\ \approx 14mS{m}^{2}mo{l}^{-1} \end{gathered}$$

Hence, The molar conductivity of this solution is $14mS{m}^{2}mo{l}^{-1}$.