Given λ°(Ca²⁺) = 119.0 and λ°(Cl⁻) = 76.3 S cm² mol⁻¹, the limiting mo…

Question

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

Accepted Answer

D. 271.6 S cm² mol⁻¹

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

Answer

195.3 S cm² mol⁻¹

Answer

152.6 S cm² mol⁻¹

Answer

313.0 S cm² mol⁻¹

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

Keep practising Unit 9B1: Electrochemical Cells and Standard Potentials