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At 298K the electrolytic conductivity of a 0.2 M KCl solution is `2.5xx10^(-2) ohm^(-1)cm^(-1)` compute its molar c onductivity.A. 62.5`ohm^(-1) cm^(2)"mol"^(-1)`B. `125ohm^(-1)cm^(2)"mol"^(-1)`C. `250ohm^(-1)cm^(2)"mol"^(-1)`D. `175ohm^(-1)cm^(2)"mol"^(-1)` |
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Answer» Correct Answer - B Molar conducitivity `=(1000K)/(C) ohm^(-1)cm^(2)"mol"^(-1)` `=(1)/(100)xx0.037=3.7xx10^(-4)ohm^(-1)cm^(-1)` `therefore `Molar conductivity `=(1000xx3.7xx10^(-4))/(0.05)` `=7.4ohm^(-1)cm^(2) "mol"^(-1)` |
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