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Question
The conductivity of 0.20 M solution of KCl at 298 K is 0.025 S cm−1. Calculate its molar conductivity.
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Solution
κ (S cm−1) = 0.025 S cm−1 and molarity (mol L−1) = 0.20 M
`Molar conductivity (A_m)=K/(1000xxmolarity)=0.025/(1000xx0.20)=1.25xx10^(-4)Scm^2mol^(-1)`
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