Question

Molar conductivities of oxalate, $K^{+}$ and ${Na}^{+}$ ions at infinite dilution are $148.2$, $50.1$, $73.5$ $S$ ${cm}^2{mol}^{-1}$ respectively. Then, equivalent conductivity at infinite dilution for sodium-potassium oxalate $\left[{\left(CO{O}^{-}\right)}_2{Na}^{+}{K}^{+}\right]$ will be:

• A. $271.8S$ ${cm}^2{eq}^{-1}$
• B. $67.95S$ ${cm}^2{eq}^{-1}$
• C. $543.6S$ ${cm}^2{eq}^{-1}$
• D. $135.9S$ ${cm}^2{eq}^{-1}$

Answer 1

Answer: (D)

$135.9S$ ${cm}^2{eq}^{-1}$

${\lambda }_m^{\infty }={\lambda }_m^{\infty }\left(oxalate\right)+{\lambda }_m^{\infty }\left({Na}^{+}\right)+{\lambda }_m^{\infty }\left(K^{+}\right)$

$=(148.2+73.5+50.1)=271.8S{cm}^2{mol}^{-1}$

$\because {\lambda }_{eq}^{\infty }=\frac{{\lambda }_m^{\infty }}{n.factor}=\frac{271.8}{2}=135.9S{cm}^2{eq}^{-1}$