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Q. $\lambda^{\circ}_m$ for $NH_4Cl, NaOH$ and $NaCl$ are $130, 248$ and $126.5\, ohm^{-1} \, cm^2 \, mol^{-1}$ respectively. The $\lambda^{\circ}_m$ of $NH_4OH$

COMEDKCOMEDK 2015Electrochemistry

Solution:

$\lambda^{\circ}_{m(NH_4OH) } = \lambda^{\circ}_{m(NH_4Cl) } + \lambda^{\circ}_{m(NaOH) } - \lambda^{\circ}_{m(NHCl) }$
$ = 130 + 248 - 126.5 = 251.5 \, ohm^{-1} \, cm^{2} \, mol^{-1} $