Answer to Question #156067 in Organic Chemistry for Akshat Kotnala

Question #156067

A conductivity cell having a cell constant of 0.5 cm–1 filled with 0.02 M solution of KCl at

298 K gave a resistance of 20.2 Ω. The water used for preparing the solution had a conductivity

of 7.1 × 10–6 S cm–1. Calculate the molar conductivity of 0.02 M KCl solution.


1
Expert's answer
2021-01-18T01:53:35-0500

Solution.

"\\lambda = \\frac{\\kappa \\times 1000}{N}"

"\\kappa = \\frac{l}{R \\times a}"

"\\kappa = \\frac{0.5}{20.2} = 0.025 \\ ohm^{-1} \\ cm^{-1}"

"\\lambda = \\frac{0.025 \\times 1000}{0.02} = 1250 \\ ohm^{-1} \\ cm^2 \\ (g-eq)^{-1}"

Answer:

"\\lambda = 1250 \\ ohm^{-1} \\ cm^2 \\ (g-eq)^{-1}"


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