The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9 S cm² mot: Calculate the conductivity of the solution.

${\mathrm{\Lambda }}_{\mathrm{m}}=138.9{\mathrm{Scm}}^2{\mathrm{mol}}^{-};$

Conductivity, $K = ?,$

Molarity of solution $=1.5\mathrm{mol}{L}^{-}.$

We know  ${\wedge }_{\mathrm{m}}=\frac{\mathrm{K}\times 100{\mathrm{cm}}^3/\mathrm{L}}{\text{ Molarity }\left({\mathrm{molL}}^{-}\right)}$

$\begin{array}{l} 138.9{\mathrm{Scm}}^2{\mathrm{mol}}^{-}=\frac{\mathrm{K}\times 1000{\mathrm{cm}}^3}{1.5{\mathrm{molL}}^{-}\times \mathrm{L}}\\ \therefore \mathrm{K}=\frac{138.9{\mathrm{Scm}}^2{\mathrm{mol}}^{-}\times 1.5\mathrm{mol}}{1000{\mathrm{cm}}^3}\\ =0.21{\mathrm{Scm}}^{-}\end{array}$