Practice question
Question
The resistance of a conductivity cell with 0.02 M KCl solution is 400 Ω, and its conductivity is 0.0025 S cm⁻¹. What is the cell constant?
Explanation
kappa = (cell constant/R) , 0.0025 = (cell constant/400) , cell constant = 1.0 cm⁻¹ .
Discussion
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