Given molar ionic conductivities $\lambda^0$: Ca²⁺ = 104, Cl⁻ = 76.4 (Ω⁻¹ cm² mol⁻¹), the molar conductivity at zero concentration of CaCl₂ is:

$\Lambda_0$(CaCl₂) = $\lambda^0$(Ca²⁺) + 2$\lambda^0$(Cl⁻) = 104 + 2(76.4) = **256.8** Ω⁻¹ cm² mol⁻¹. The factor of 2 accounts for two chloride ions per formula unit.