If λm∞12Zn2+=52.8Scm2mol−1, and Λm∞Zn3Fe(CN)62=922.8Scm2mol−1 The Λm∞Fe(CN)63− will be

A
$101 S {cm}^{2} {mol}^{- 1}$
B
$202 S {cm}^{2} {mol}^{- 1}$
C
$303 S {cm}^{2} {mol}^{- 1}$
D
$404 S {cm}^{2} {mol}^{- 1}$

Solution:

$Λ_{m} ({Zn}_{3} {[Fe (CN)_{6}]}_{2}) = 3 λ_{m} ({Zn}^{2 +}) + 2 λ_{m} (Fe (CN)_{6}^{3 -})$ $= 3 \times 2 λ_{m} (\frac{1}{2} {Zn}^{2 +}) + 2 λ_{m} (Fe (CN)_{6}^{3 -})$ or $λ_{m} (Fe (CN)_{6}^{3 -}) = \frac{1}{2} [Λ_{m} ({Zn}_{3} {[Fe (CN)_{6}]}_{2}) - 6 λ_{m} (\frac{1}{2} {Zn}^{2 +})] = \frac{1}{2} [922.8 - 6 \times 52.8] {Scm}^{2} {mol}^{- 1} = 303 S {cm}^{2} {mol}^{- 1}$

If $λ_{m}^{\infty} (\frac{1}{2} {Zn}^{2 +}) = 52.8 S {cm}^{2} {mol}^{- 1}$ , and $Λ_{m}^{\infty} ({Zn}_{3} {[Fe (CN)_{6}]}_{2}) = 922.8 {Scm}^{2} {mol}^{- 1}$ The $Λ_{m}^{\infty} [Fe (CN)_{6}^{3 -}]$ will be

Solution:

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