Question

Given the limiting molar conductivity as:

$A_m^{\infty }\left(HCl\right)=126.4{\Omega }^{-1}{cm}^2{mol}^{-1}$
$A_m^{\infty }\left(NaCl\right)=425.9{\Omega }^{-1}{cm}^2{mol}^{-1}$
$A_m^{\infty }\left({CH}_{3}COONa\right)=91{\Omega }^{-1}{cm}^2{mol}^{-1}$

The molar conductivity at infinite dilution of acetic acid $\left(in{\Omega }^{-1}{cm}^2{mol}^{-1}\right)$ will be:

• A. $481.5$
• B. $390.5$
• C. $299.5$
• D. $516.9$

Answer 1

Answer: (B)

$390.5$

According to Kohlraush's law: Each ion makes a definite contribution to the total molar conductivity of an electrolyte irrespective of the nature of the other ion.

${\Lambda }_m^o=X_{\lambda \left(A^{y+}\right)}^o+Y_{\lambda \left(B^{x-}\right)}$

Similarly,

${\Lambda }_{m\left({CH}_{3}COOH\right)}^{\infty }={\Lambda }_{m\left({CH}_{3}COONa\right)}^{\infty }+{\Lambda }_{m\left(HCl\right)}^{\infty }-{\Lambda }_{m\left(NaCl\right)}^{\infty }$

${\Lambda }_{m\left({CH}_{3}COOH\right)}^{\infty }=91+425.9-126.4=390.5{\Omega }^{-1}{cm}^2{mol}^{-1}$

Hence, option B is correct.