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The molar conductivity of 0.05 M `BaCI_(2)` solution at `25^(@)C ` is `223 Omega^(-1) cm^(2) mol^(-1)` .What is its conductivity ? |
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Answer» Correct Answer - Conductivity = k = 0.01115 `Omega^(-1) cm^(-1)` Molar conductivity `=^^_(m) =223 Omega^(-1) cm^(2) mol^(-1)` Concetration = C= 0.05 M `BaCI_(2)` Conductivity = k = ? `^^_(m) = (k xx 1000)/(C )` `:. K = (^^_(m) xx C)/(1000) = (223 xx 0.05)/(1000) = 0.01115 Omega^(-1) cm^(-1)` |
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