Electrochemistry Ques 35
35. Limiting molar conductivity of $\mathrm{NH} _4 \mathrm{OH}$ (i.e., $\left.\Lambda _{\mathrm{m}\left(\mathrm{NH} _4 \mathrm{OH}\right)}^{\circ}\right)$ is equal to :
[2012]
(a) $\Lambda_ {\mathrm{m}\left(\mathrm{NH}_ 4 \mathrm{Cl}\right)}^{\circ}+\Lambda _{\mathrm{m}(\mathrm{NaCl})}^\circ-\Lambda _{\mathrm{m}(\mathrm{NaOH})}^{\circ}$
(b) $\Lambda_{\mathrm{m}(\mathrm{NaOH})}^{\circ}+\Lambda_{\mathrm{m}(\mathrm{NaCl})}^{\circ}-\Lambda_{\mathrm{m}\left(\mathrm{NH}_4 \mathrm{Cl}\right)}^{\circ}$
(c) $\left.\Lambda _{\mathrm{m}}^{\circ}\left(\mathrm{NH}_4 \mathrm{OH}\right)+\Lambda_0^{\circ} \mathrm{mNH} _4 \mathrm{Cl}\right)-\Lambda _{\mathrm{m}(\mathrm{HCl})}^{\circ}$
(d) $\Lambda_{\mathrm{m}\left(\mathrm{NH}_ 4 \mathrm{Cl}\right)}^{\circ}+\Lambda _{\mathrm{m}(\mathrm{NaOH})}^{\circ}-\Lambda _{\mathrm{m}(\mathrm{NaCl})}^{\circ}$
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Answer:
Correct Answer: 35.(d)
Solution:
BETA
