The equation that is incorrect is

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(A) According to Kohlrausch's law,the molar conductivity at infinite dilution is the sum of the contributions of individual ions.For option $A$: $(\La

Vedclass · Accepted Answer

$(\Lambda_{m}^{0})_{NaBr} - (\Lambda_{m}^{0})_{NaI} = (\Lambda_{m}^{0})_{KBr} - (\Lambda_{m}^{0})_{NaBr}$

Vedclass · Answer

(A) According to Kohlrausch's law,the molar conductivity at infinite dilution is the sum of the contributions of individual ions.For option $A$: $(\Lambda_{m}^{0})_{NaBr} - (\Lambda_{m}^{0})_{NaI} = (\lambda_{Na^+} + \lambda_{Br^-}) - (\lambda_{Na^+} + \lambda_{I^-}) = \lambda_{Br^-} - \lambda_{I^-}$.On the right side: $(\Lambda_{m}^{0})_{KBr} - (\Lambda_{m}^{0})_{NaBr} = (\lambda_{K^+} + \lambda_{Br^-}) - (\lambda_{Na^+} + \lambda_{Br^-}) = \lambda_{K^+} - \lambda_{Na^+}$.Since $\lambda_{Br^-} - \lambda_{I^-} 
eq \lambda_{K^+} - \lambda_{Na^+}$,this equation is incorrect.For option $B$: $(\Lambda_{m}^{0})_{HCl} + (\Lambda_{m}^{0})_{NaOH} - (\Lambda_{m}^{0})_{NaCl} = (\lambda_{H^+} + \lambda_{Cl^-}) + (\lambda_{Na^+} + \lambda_{OH^-}) - (\lambda_{Na^+} + \lambda_{Cl^-}) = \lambda_{H^+} + \lambda_{OH^-} = (\Lambda_{m}^{0})_{H_{2}O}$. This is correct.For option $C$: $(\Lambda_{m}^{0})_{KCl} - (\Lambda_{m}^{0})_{NaCl} = \lambda_{K^+} - \lambda_{Na^+}$ and $(\Lambda_{m}^{0})_{KBr} - (\Lambda_{m}^{0})_{NaBr} = \lambda_{K^+} - \lambda_{Na^+}$. This is correct.For option $D$: $(\Lambda_{m}^{0})_{NaBr} - (\Lambda_{m}^{0})_{NaCl} = \lambda_{Br^-} - \lambda_{Cl^-}$ and $(\Lambda_{m}^{0})_{KBr} - (\Lambda_{m}^{0})_{KCl} = \lambda_{Br^-} - \lambda_{Cl^-}$. This is correct.

List-$I$ (Conversion)	List-$II$ (Number of Faraday required)
$A$. $1 \ mol$ of $H_2O$ to $O_2$	$I$. $3 \ F$
$B$. $1 \ mol$ of $MnO_4^{-}$ to $Mn^{2+}$	$II$. $2 \ F$
$C$. $1.5 \ mol$ of $Ca$ from molten $CaCl_2$	$III$. $1 \ F$
$D$. $1 \ mol$ of $FeO$ to $Fe_2O_3$	$IV$. $5 \ F$

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