Question

Molar conductives $\left({\Lambda }_m^o\right)$ at infinite dilution of $NaCl,HCl$ and ${CH}_{3}COONa$ are $126.4,425.9$ and $91.0S{cm}^2{mol}^{-1}$ respectively. ${\Lambda }_m^o$ for ${CH}_{3}COOH$ will be:

• A. $180.5S{cm}^2{mol}^{-1}$
• B. $290.8S{cm}^2{mol}^{-1}$
• C. $390.5S{cm}^2{mol}^{-1}$
• D. $425.5S{cm}^2{mol}^{-1}$

Answer 1

Answer: (C)

$390.5S{cm}^2{mol}^{-1}$

Let x be the molar conductivity of acetic acid.

${\Lambda }_{m\left({CH}_{3}COONa\right)}^o+{\Lambda }_{m\left(HCl\right)}^o\to {\Lambda }_{m\left(NaCl\right)}^o+{\Lambda }_{m\left({CH}_{3}COOH\right)}^o$

$91+425.9=126.4+x$

$\therefore$ $x=390.5$ ${ohm}^{-1}{cm}^2{mol}^{-1}$

Thus, the molar conductivity of acetic acid is $390.5S{cm}^2{mol}^{-1}$.