Question

Calculate the molar conductance $\Lambda_m^\circ$ for CaCl$_2$, given that $\lambda^\circ(Ca^{+2}) = 119.5$ S cm$^2$ mol$^{-1}$ and $\lambda^\circ(Cl^-) = 76.3$ S cm$^2$ mol$^{-1}$.

Hint:

Don't forget the stoichiometry! CaCl$_2$ produces two chloride ions, so you must multiply the chloride conductivity by 2.

Answer 1

According to Kohlrausch's Law of independent migration of ions, the limiting molar conductivity of an electrolyte is the sum of the limiting ionic conductivities of the cation and anion, multiplied by the number of ions per formula unit.
Formula for CaCl$_2$:
$\Lambda_m^\circ(\text{CaCl}_2) = \lambda^\circ(\text{Ca}^{2+}) + 2 \times \lambda^\circ(\text{Cl}^-)$
Substitute values:
$\Lambda_m^\circ = 119.5 + 2 \times (76.3)$
$\Lambda_m^\circ = 119.5 + 152.6$
$\Lambda_m^\circ = 272.1$ S cm$^2$ mol$^{-1}$.