Question

Molar conductivities $\left({\wedge }_m\right)$ at infinite dilution of NaCl, HCl and $C{H}_{3}COONa$ are 126.4, 425.9 and 91.0 S $c{m}^{2}mo{l}^{-1}$ respectively. $\left({\wedge }_m\right)$ for $C{H}_{3}COOH$ will be:

• A. $180.5Sc{m}^{2}mo{l}^{-1}$
• B. $290.8Sc{m}^{2}mo{l}^{-1}$
• C. $390.5Sc{m}^{2}mo{l}^{-1}$
• D. $425.5Sc{m}^{2}mo{l}^{-1}$

Answer 1

Answer: (C)

$390.5Sc{m}^{2}mo{l}^{-1}$

Kohlrausch law: It states that the molar conductivity of an electrolyte at infinite dilution can be expressed as the sum of the contributions from its individual ions.

${\Lambda }_{N{a}^{+}}=x{\Lambda }_{H^{+}}=y{\Lambda }_{C{l}^{-}}=z{\Lambda }_{C{H}_{3}CO{O}^{-}}=w$

$x+z=126.4$

$y+z=425.9$

$w+x=91$

from the above equation, value of

$w+y=390.5$