Question

Equivalent conductivity at infinite dilution for sodium-potassium oxalate $((COO^-)_2Na^+K^+)$ will be [given, molar conductivities of oxalate, $K^+$ and $Na^+$ ions at infinite dilution are $148.2, 50.1, 73.5 \,S$ $cm^2mol^{-1}$, respectively]

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

Answer 1

Answer: (D)

$\lambda_{m}^{\infty}=\lambda_{m}^{\infty}$ (oxalate $)+\lambda_{m}^{\infty}\left( Na ^{+}\right)+\lambda_{m}^{\infty}\left( K ^{+}\right)$
$=(148 \cdot 2+73 \cdot 5+50 \cdot 1)$
$=271.85 \,cm ^{2}\, mol ^{-1}$
$ \because \lambda_{ eq }^{\infty} =\frac{\lambda_{m}^{\infty}}{n \cdot \text { factor }}=\frac{271.8}{2} $
$ =135.9 \,S \,cm ^{2} eq ^{-1}$