Question
How is the unit of molar conductivity arrived at?
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Answer
$\Lambda_{\text{m}}=\text{k}\times\text{V}=\text{ohm}^{-1}\text{cm}^{-1}\times\text{cm}^3\text{mol}^{-1}$
$=\text{ohm}^{-1}\text{cm}^{2}\text{mol}^{-1}=\text{S cm}^2\text{mol}^{-1}$
$=\text{ohm}^{-1}\text{cm}^{2}\text{mol}^{-1}=\text{S cm}^2\text{mol}^{-1}$
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