Given λ°(Ca²⁺) = 119.0 and λ°(Cl⁻) = 76.3 S cm² mol⁻¹, the limiting molar conductivity of CaCl₂ is:

CaCl₂ → Ca²⁺ + 2Cl⁻. So Λ°_m = λ°(Ca²⁺) + 2 × λ°(Cl⁻) = 119.0 + 2(76.3) = 119.0 + 152.6 = **271.6 S cm² mol⁻¹**.