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26. The limiting molar conductivities () for , , and are 126, 152, and 150 respectively. Using KohlrElectrochemistry Chemistry Question

Question

26. The limiting molar conductivities ($\Lambda^\circ_m$) for $NaCl$, $KBr$, and $KCl$ are 126, 152, and 150 $S \text{ cm}^2 \text{ mol}^{-1}$ respectively. Using Kohlrausch's law, what is the exact $\Lambda^\circ_m$ for $NaBr$?

Answer: A

💡 Solution & Explanation

By Kohlrausch's law: $\Lambda^\circ_m(NaBr) = \Lambda^\circ_m(NaCl) + \Lambda^\circ_m(KBr) - \Lambda^\circ_m(KCl) = 126 + 152 - 150 = 278 - 150 = 128 \text{ S cm}^2 \text{ mol}^{-1}$.

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