Chemistry - Electrochemistry Question with Solution | TestHub

The specific conductance $\kappa$ is given by $\kappa ={\Lambda }_m\cdot C$, where ${\Lambda }_m$ is molar conductivity. At infinite dilution ($C\to 0$), ${\Lambda }_m={\Lambda }_m^{\circ }$.

For NaCl: ${\kappa }_{NaCl}=({\lambda }^{\circ }({\mathrm{Na}}^{+})+{\lambda }^{\circ }({\mathrm{Cl}}^{-}))\cdot C$.

For MgSO₄: ${\kappa }_{MgS{O}_4}=({\lambda }^{\circ }({\mathrm{Mg}}^{2+})+{\lambda }^{\circ }({\mathrm{SO}}_4^{2-}))\cdot C$.

The ratio is $\frac{{\kappa }_{MgS{O}_4}}{{\kappa }_{NaCl}}=\frac{{\lambda }^{\circ }({\mathrm{Mg}}^{2+})+{\lambda }^{\circ }({\mathrm{SO}}_4^{2-})}{{\lambda }^{\circ }({\mathrm{Na}}^{+})+{\lambda }^{\circ }({\mathrm{Cl}}^{-})}$.