A $5.0 \mathrm{m} \mathrm{mol} {\mathrm{dm}}^{-3}$ aqueous solution of $\mathrm{KCl}$ has a conductance of $0.55 \mathrm{mS}$ when measured in a cell constant $1.3{ \mathrm{cm}}^{-1}$. The molar conductivity of this solution is _______ ${\mathrm{mSm}}^2{ \mathrm{mol}}^{-1}$. (Round off to the Nearest Integer)

Given ${\mathrm{conc}}^{\mathrm{n}}$ of $\mathrm{KCl}=\frac{\mathrm{m}.\mathrm{mol}}{\mathrm{L}}$

: Conductance $\left(\mathrm{G}\right)=0.55 \mathrm{mS}$

: Cell constant $\left(\frac{\ell }{\mathrm{A}}\right)=1.3{ \mathrm{cm}}^{-1}$

To Calculate : Molar conductivity $\left({\lambda }_{\mathrm{m}}\right)$ of sol.

$\to$ Since ${\mathrm{\Lambda }}_{\mathrm{m}}=\frac{1}{1000}\times \frac{\mathrm{k}}{\mathrm{M}}$  $...\left(1\right)$

$\to$ Molarity $=5\times {10}^{-3}\frac{\mathrm{mol}}{\mathrm{L}}$

$\to$ Conductivity $=\mathrm{G}\times \left(\frac{\ell }{\mathrm{A}}\right)=0.55 \mathrm{mS}\times \frac{1.3}{\frac{1}{100}}{ \mathrm{m}}^{-1}$

$=55\times 1.3 {\mathrm{mSm}}^{-1}$

${\mathrm{eq}}^{\mathrm{n}} \left(1\right)$ ${\mathrm{\Lambda }}_{\mathrm{m}}=\frac{1}{1000}\times \frac{55\times 1.3}{\left(\frac{5}{1000}\right)}\frac{{\mathrm{mSm}}^2}{ \mathrm{mol}}$

$\Rightarrow {\mathrm{\Lambda }}_{\mathrm{m}}=14.3\frac{{\mathrm{mSm}}^2}{ \mathrm{mol}}$