Practice question
Question
The conductivity of a 0.025 M NaNO₃ solution is 0.003 S cm⁻¹, and its molar conductivity is 120 S cm² mol⁻¹. What is the degree of dissociation if Lambdam° = 126.5 S cm² mol⁻¹ ?
Explanation
α = (Lambdam/Lambdam°) = (120/126.5) = 0.9486 ≈ 0.95 .
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