During the electrolysis of carnallite,$MgCl_2$ is decomposed and not $KCl$. This is because of

  • A
    lower decomposition voltage of $MgCl_2$ than that of $KCl$
  • B
    reverse reaction $MgCl_2 + 2K \to Mg + 2KCl$ if $KCl$ is decomposed under other experimental conditions
  • C
    both $(A)$ and $(B)$
  • D
    none of the above

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If $\Lambda^{0}_{NaCl} = 126 \ S \ cm^{2} \ mol^{-1}$,$\Lambda^{0}_{KBr} = 125 \ S \ cm^{2} \ mol^{-1}$,and $\Lambda^{0}_{KCl} = 150 \ S \ cm^{2} \ mol^{-1}$,then find $\Lambda^{0}_{NaBr}$ in $S \ cm^{2} \ mol^{-1}$.

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