Q77 / 200 · NEET 2012 (Screening) · Chemistry

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

  • A
    $\stackrel{\circ}{\Lambda}_{m\left(\mathrm{NH}_4 \mathrm{Cl}\right)}+\stackrel{\circ}{\Lambda}_{m(\mathrm{NaCl})}-\stackrel{\circ}{\Lambda}_{m(\mathrm{NaOH})}$
  • B
    $\stackrel{\circ}{\Lambda}_{m(\mathrm{NaOH})}+\stackrel{\circ}{\Lambda}_{m(\mathrm{NaCl})}-\stackrel{\circ}{\Lambda}_{m\left(\mathrm{NH}_4 \mathrm{Cl}\right)}$
  • C
    $\stackrel{\circ}{\Lambda}_{m\left(\mathrm{NH}_4 \mathrm{OH}\right)}+\stackrel{\circ}{\Lambda}_{m\left(\mathrm{NH}_4 \mathrm{Cl}\right)}-\stackrel{\circ}{\Lambda}_{m(\mathrm{HCl})}$
  • Correct answer
    D
    $\stackrel{\circ}{\Lambda}_{m\left(\mathrm{NH}_4 \mathrm{Cl}\right)}+\stackrel{\circ}{\Lambda}_{m(\mathrm{NaOH})}-\stackrel{\circ}{\Lambda}_{m(\mathrm{NaCl})}$

Solution

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

Hence, the correct answer is option: 'D'.

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