COMEDK · Chemistry · 18. Electrochemistry
\(\lambda_{\mathrm{m}}^{\circ}\) for \(\mathrm{NH}_{4} \mathrm{Cl}, \mathrm{NaOH}\) and \(\mathrm{NaCl}\) are 130,248 and \(126.5 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\) respectively. The \(\lambda_{\mathrm{m}}^{\circ}\) of \(\mathrm{NH}_{4} \mathrm{OH}\) will be
- A \(251.5\)
- B \(244.5\)
- C 130
- D \(504.5\)
Answer & Solution
Correct Answer
(A) \(251.5\)
Step-by-step Solution
Detailed explanation
\(\lambda_{\mathrm{m}\left(\mathrm{NH}_{4} \mathrm{OH}\right)}^{\circ}=\lambda_{\mathrm{m}\left(\mathrm{NH}_{4} \mathrm{Cl}\right)}^{\circ}+\lambda_{\mathrm{m}(\mathrm{NaOH})}^{\circ}-\lambda_{\mathrm{m}(\mathrm{NaCl})}^{\circ}\)…
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