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Limiting molar conductivity of $\mathrm{NH}_4 \mathrm{OH~} \left( \text{i.e., } \stackrel{\circ}{\Lambda}_{m\left(\mathrm{NH}_4 \mathrm{OH}\right)}\right)$ is equal to
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The correct answer is:
$\stackrel{\circ}{\Lambda}_{m\left(\mathrm{NH}_4 \mathrm{Cl}\right)}+\stackrel{\circ}{\Lambda}_{m(\mathrm{NaOH})}-\stackrel{\circ}{\Lambda}_{m(\mathrm{NaCl})}$
According to Kohlrausch's law, limiting molar conductivity of $\mathrm{NH}_4 \mathrm{OH}$
$$
\Lambda_{m\left(\mathrm{NH}_4 \mathrm{OH}\right)}^{\circ}=\Lambda_{m\left(\mathrm{NH}_4 \mathrm{Cl}\right)}^{\circ}+\Lambda_{m(\mathrm{NaOH})}^{\circ}-\Lambda_{m(\mathrm{NaCl})}^{\circ}
$$
$$
\Lambda_{m\left(\mathrm{NH}_4 \mathrm{OH}\right)}^{\circ}=\Lambda_{m\left(\mathrm{NH}_4 \mathrm{Cl}\right)}^{\circ}+\Lambda_{m(\mathrm{NaOH})}^{\circ}-\Lambda_{m(\mathrm{NaCl})}^{\circ}
$$
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