Chemistry - Electrochemistry Question with Solution | TestHub

Using the formula,

$\mathrm{\kappa } \left(\mathrm{conductivity}\right)=\frac{\mathrm{Cell}\mathrm{ }\mathrm{constant}}{\mathrm{R}}=\frac{1.15}{250} \mathrm{S} {\mathrm{cm}}^{-1}$

Now, using the relation between molar conductance, conductivity and molarity of the solution,

$$\begin{gathered} {\mathrm{\Lambda }}_{\mathrm{m}} \left(\mathrm{S} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1}\right)=\frac{\mathrm{K} \left(\mathrm{S} {\mathrm{cm}}^{-1}\right)\times 1000 \left({\mathrm{cm}}^3/\mathrm{L}\right)}{\mathrm{molarity} (\mathrm{mol}/\mathrm{L})} \\ {\mathrm{\Lambda }}_{\mathrm{m}}=\frac{1.15}{250}\times \frac{1000}{1} \\ {\mathrm{\Lambda }}_{\mathrm{m}}=4.6\mathrm{ }\mathrm{S} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1}. \end{gathered}$$