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The limiting equivalent conductivity of `NaCl, KCl` and `KBr` are `126.5,150.0` and `151.5 S cm^(2) eq^(-1)` , respectively. The limiting equivalent ionic conductance for `Br^(-)` is `78 S cm^(2) eq^(-1)`. The limiting equivalent ionic conductance for `Na^(+)` ions would be `:`A. 128B. 125C. 49D. 50 |
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Answer» Correct Answer - 4 `Lambda_(m)^(oo)=(NaBr)=Lambda_(m)^(oo)(NaCl)+Lambda_(m)^(oo)(KBr)-Lambda_(m)^(oo)(KCl)` `Lambda_(m)^(oo)(Na^(+))+Lambda_(m)^(oo)(Br)=126.5+151.5-150` `Lambda_(m)^(oo)(Na^(+))=50` |
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