Question

Given the limiting molar conductivity as
$\hat{}0m\left(HCl\right)=425.9\Omega -1cm2mol-1$
$\hat{}0m\left(NaCl\right)=126.4\Omega -1cm2mol-1$
$\hat{}0m\left(CH3COONa\right)=91\Omega -1cm2mol-1$
The molar conductivity, at infinite dilution, of acetic (in $\Omega -1cm2mol-1$) will be:

• A. 481.5
• B. 390.5390.
• C. $299.5$
• D. $516.9$

Answer 1

Answer: (B)

390.5390.

According to Kphlrush's law. Each ion makes a definite contribution to the total molar conductivity of an electrode irrespective of the nature of the other ion
eg for $A_x{B}_y$
$S_m^o=X_{\lambda \left(A^{y+}\right)}^o+Y_{\lambda \left(B^{x-}\right)}$
similarly
${\Lambda }_{m\left(C{H}_{2}COOH\right)}^{\infty }={\Lambda }_{m\left(C{H}_{2}COOH\right)}^{\infty }+{\Lambda }_{m\left(HCl\right)}^{\infty }-{\Lambda }_{m\left(NaCl\right)}^{\infty }$
$=91+425.9-126.4=390{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$