The limiting molar conductivity $\Lambda^o$ for $NaCl$,$KBr$ and $KCl$ are $126$,$152$ and $150 \ S \ cm^2 \ mol^{-1}$ respectively. The $\Lambda^o$ for $NaBr$ is (in $S \ cm^2 \ mol^{-1}$).

  • A
    $302$
  • B
    $176$
  • C
    $278$
  • D
    $128$

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If $\lambda_{m}^{o}(Br^{-}) = 78.1 \, S \, cm^{2} \, mol^{-1}$,calculate $\lambda_{m}^{o}(K^{+})$ using the following data:
Electrolyte $\Lambda_{m}^{o} \, (S \, cm^{2} \, mol^{-1})$
$NaCl$ $126.5$
$NaBr$ $128.2$
$KCl$ $149.5$

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