If lambda_{m}^{o}(Br^{-}) = 78.1 , S , cm^{2} | vedclass

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(B) According to Kohlrausch's law,the molar conductivity of an electrolyte at infinite dilution is the sum of the ionic conductivities of its constitu

Vedclass · Accepted Answer

$73.5 \, S \, cm^{2} \, mol^{-1}$

Vedclass · Answer

(B) According to Kohlrausch's law,the molar conductivity of an electrolyte at infinite dilution is the sum of the ionic conductivities of its constituent ions.$1$. For $NaBr$: $\Lambda_{m}^{o}(NaBr) = \lambda_{m}^{o}(Na^{+}) + \lambda_{m}^{o}(Br^{-}) = 128.2 \, S \, cm^{2} \, mol^{-1}$.Given $\lambda_{m}^{o}(Br^{-}) = 78.1 \, S \, cm^{2} \, mol^{-1}$,we find $\lambda_{m}^{o}(Na^{+}) = 128.2 - 78.1 = 50.1 \, S \, cm^{2} \, mol^{-1}$.$2$. For $NaCl$: $\Lambda_{m}^{o}(NaCl) = \lambda_{m}^{o}(Na^{+}) + \lambda_{m}^{o}(Cl^{-}) = 126.5 \, S \, cm^{2} \, mol^{-1}$.Substituting $\lambda_{m}^{o}(Na^{+}) = 50.1$,we get $\lambda_{m}^{o}(Cl^{-}) = 126.5 - 50.1 = 76.4 \, S \, cm^{2} \, mol^{-1}$.$3$. For $KCl$: $\Lambda_{m}^{o}(KCl) = \lambda_{m}^{o}(K^{+}) + \lambda_{m}^{o}(Cl^{-}) = 149.5 \, S \, cm^{2} \, mol^{-1}$.Substituting $\lambda_{m}^{o}(Cl^{-}) = 76.4$,we get $\lambda_{m}^{o}(K^{+}) = 149.5 - 76.4 = 73.1 \, S \, cm^{2} \, mol^{-1}$.Alternatively,$\lambda_{m}^{o}(K^{+}) = \Lambda_{m}^{o}(KCl) - \Lambda_{m}^{o}(NaCl) + \Lambda_{m}^{o}(NaBr) - \lambda_{m}^{o}(Br^{-}) = 149.5 - 126.5 + 128.2 - 78.1 = 73.1 \, S \, cm^{2} \, mol^{-1}$.Rounding to the nearest option,the answer is $73.5 \, S \, cm^{2} \, mol^{-1}$.

Electrolyte	$\Lambda_{m}^{o} \, (S \, cm^{2} \, mol^{-1})$
$NaCl$	$126.5$
$NaBr$	$128.2$
$KCl$	$149.5$

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