Limiting molar conductivity of $\mathrm{NH}_4 \mathrm{OH~} \left( \text{i.e., }…

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
  1. $\stackrel{\circ}{\Lambda}_{m\left(\mathrm{NH}_4 \mathrm{Cl}\right)}+\stackrel{\circ}{\Lambda}_{m(\mathrm{NaCl})}-\stackrel{\circ}{\Lambda}_{m(\mathrm{NaOH})}$
  2. $\stackrel{\circ}{\Lambda}_{m(\mathrm{NaOH})}+\stackrel{\circ}{\Lambda}_{m(\mathrm{NaCl})}-\stackrel{\circ}{\Lambda}_{m\left(\mathrm{NH}_4 \mathrm{Cl}\right)}$
  3. $\stackrel{\circ}{\Lambda}_{m\left(\mathrm{NH}_4 \mathrm{OH}\right)}+\stackrel{\circ}{\Lambda}_{m\left(\mathrm{NH}_4 \mathrm{Cl}\right)}-\stackrel{\circ}{\Lambda}_{m(\mathrm{HCl})}$
  4. $\stackrel{\circ}{\Lambda}_{m\left(\mathrm{NH}_4 \mathrm{Cl}\right)}+\stackrel{\circ}{\Lambda}_{m(\mathrm{NaOH})}-\stackrel{\circ}{\Lambda}_{m(\mathrm{NaCl})}$

Solution

According to Kohlrausch's law, limiting molar conductivity of $\mathrm{NH}_4 \mathrm{OH}$ $\Lambda_{m\left(\mathrm{NH}_4 \mathrm{OH}\right)}^{\circ}=\Lambda_{m\left(\mathrm{NH}_4 \mathrm{Cl}\right)}^{\circ}+\Lambda_{m(\mathrm{NaOH})}^{\circ}-\Lambda_{m(\mathrm{NaCl})}^{\circ}$

Asked in: NEET 2012 (Screening)

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