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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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Answer» Solution :Given that k = `0.025 S cm^(-1)` `C = 0.2 "moles" L^(-1) = 0.2` moles `(1000 cm^(3))^(-1) =(0.2 "moles")/(1000 cm^(3))` `Lambda_(m) = k/C = (0.025 S cm^(-1))/(0.2 "moles") XX 1000 cm^(3) = 125 S cm^(2) "mol"^(-1)` |
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