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The limmiting molar conductivites `^^ ^(@)` for `NaCl_(2) ` Kbr and KCl are 126,152 and 150 S `cm^(2)` respectively the `^^ ^(@)` for NaBr isA. `128Scm^(2)"mol"^(-1)`B. `302Scm^(2)"mol"^(-1)`C. `278S cm^(2)"mol"^(-1)`D. `176Scm^(2)"mol"^(-1)` |
Answer» Correct Answer - A `^^_(NaBr)^(@)=^^_(NaCl)^(@)+^^_(KBr)^(@)-^^_(KCl)^(@)` `=126+152-150=128 S cm^(2)"mol"^(-1)` |
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