Practice question
Question
The limiting molar conductivity of CaCl₂ is 258 S cm² mol⁻¹. If λ°Ca²⁺ = 119 S cm² mol⁻¹ , what is λ°Cl⁻ ?
Explanation
Lambdam° = λ°Ca²⁺ + 2λ°Cl⁻ . 258 = 119 + 2λ°Cl⁻ , 2λ°Cl⁻ = 139 , λ°Cl⁻ = 69.5 S cm² mol⁻¹ .
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