The value of $\Lambda_{eq}^{\infty}$ for $NH_{4}Cl,\ NaOH\ and\ NaCl$ are respectively,149.74,248.1 and 126.4$\mathrm{\Omega}^{- 1}\ {cm}^{2}equiv^{- 1}$. The value of $\Lambda_{eq}^{\infty}\$of $NH_{4}OH$ is

Question

Dr. Abubacker · Accepted Answer

(b)
$\Lambda_{eq}^{\infty}\left( NH_{4}\ OH \right) = \Lambda_{eq}^{\infty}\ \left( NH_{4}Cl \right) + \Lambda_{eq}^{\infty}\ (NaOH) - \Lambda_{eq}^{\infty}\ (NaCl)$
= (149.74 + 248.1 – 126.4)
$\ \ \ \ \ \ \ \ \  = \ 271.44\ \mathrm{\Omega}^{- 1}\ {cm}^{2}{eq}^{- 1}$

Q. The value of $\Lambda_{eq}^{\infty}$ for $NH_{4}Cl,\ NaOH\ and\ NaCl$ are respectively,149.74,248.1 and 126.4$\mathrm{\Omega}^{- 1}\ {cm}^{2}equiv^{- 1}$. The value of $\Lambda_{eq}^{\infty}\$of $NH_{4}OH$ is

Explanation

Reviewed by Dr. Abubacker

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