If lambda_{m}^{o}(SO_{4}^{2-}) = 160 S cm^{ | vedclass

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(A) According to Kohlrausch's law of independent migration of ions:$Al_{2}(SO_{4})_{3} \rightarrow 2 Al^{3+}_{(aq)} + 3 S{O_{4}}^{2-}_{(aq)}$$\Lambda_

Vedclass · Accepted Answer

$139 \ S \ cm^{2} \ mol^{-1}$

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(A) According to Kohlrausch's law of independent migration of ions:$Al_{2}(SO_{4})_{3} \rightarrow 2 Al^{3+}_{(aq)} + 3 S{O_{4}}^{2-}_{(aq)}$$\Lambda_{m}^{o}[Al_{2}(SO_{4})_{3}] = 2 \lambda_{m}^{o}(Al^{3+}) + 3 \lambda_{m}^{o}(SO_{4}^{2-})$Given: $\Lambda_{m}^{o}[Al_{2}(SO_{4})_{3}] = 758 \ S \ cm^{2} \ mol^{-1}$ and $\lambda_{m}^{o}(SO_{4}^{2-}) = 160 \ S \ cm^{2} \ mol^{-1}$Let $\lambda_{m}^{o}(Al^{3+}) = x$$758 = 2(x) + 3(160)$$758 = 2x + 480$$2x = 758 - 480 = 278$$x = \frac{278}{2} = 139 \ S \ cm^{2} \ mol^{-1}$

If $\lambda_{m}^{o}(SO_{4}^{2-}) = 160 \ S \ cm^{2} \ mol^{-1}$ and $\Lambda_{m}^{o}(Al_{2}(SO_{4})_{3}) = 758 \ S \ cm^{2} \ mol^{-1}$,then calculate $\lambda_{m}^{o}(Al^{3+})$.

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