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Molar conductivities ($\Lambda_m^0$) at infinite dilution of NaCl, HCl and $CH_3COONa$ are 126.4, 425.9 and 91.0 S $cm^2$ $mol^{-1}$ respectively. $\Lambda_m^0$ for $CH_3COOH$ will be:
A
390.5 S $cm^2$ $mol^{-1}$
B
425.5 S $cm^2$ $mol^{-1}$
C
180.5 S $cm^2$ $mol^{-1}$
D
290.8 S $cm^2$ $mol^{-1}$
Detailed Solution
By Kohlrausch's law: $\Lambda_m^0(CH_3COOH) = \Lambda_m^0(CH_3COONa) + \Lambda_m^0(HCl) - \Lambda_m^0(NaCl)$
(The $Na^+$ and $Cl^-$ contributions cancel, leaving $CH_3COO^-$ and $H^+$.)
$= 91.0 + 425.9 - 126.4 = 390.5$ S $cm^2$ $mol^{-1}$
(The $Na^+$ and $Cl^-$ contributions cancel, leaving $CH_3COO^-$ and $H^+$.)
$= 91.0 + 425.9 - 126.4 = 390.5$ S $cm^2$ $mol^{-1}$