Electrochemistry - Result Question 12
####13. Equivalent conductances of $NaCl, HCl$ and $CH_3 COONa$ at infinite dilution are 126.45,426.16 and $91 ohm^{-1} cm^{2}$ respectively. The equivalent conductance of $CH_3 COOH$ at infinite dilution would be
[1997]
(a) $101.38 ohm^{-1} cm^{2}$
(b) $253.62 ohm^{-1} cm^{2}$
(c) $390.71 ohm^{-1} cm^{2}$
(d) $678.90 ohm^{-1} cm^{2}$
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Solution:
- (c) By Kohlraush’s law, $\wedge_eq^{\circ} NaCl=126.45$
$\lambda _{Na^{+}}^{\circ}+\lambda _{Cl^{-}}^{\circ}=126.45$
$\lambda _{H^{+}}^{\circ}+\lambda _{Cl^{-}}^{\circ}=426.16$
$\lambda _{CH_3 OO^{-}}^{\circ}+\lambda _{Na^{+}}^{\circ}=91$
on adding (2) and (3) then subtract (1) from it
$\lambda _{CH_3 COO^{-}}^{\circ}+\lambda _{H^{+}}^{\circ}=517.16-126.45$
$\graveU _{(CH_3 COOH)}^{\circ}=390.71 ohm^{-1} cm^{2}$
