Question

Equivalent conductivity at infinite dilution for sodium-potassium oxalate $((CO{O}^{-}{)}_{2}N{a}^{+}{K}^{+})$ will be [given, molar conductivities of oxalate, $K^{+}$ and $N{a}^{+}$ ions at infinite dilution are $148.2,50.1,73.5S$ $c{m}^{2}mo{l}^{-1}$, respectively]

• A. $271.8Sc{m}^{2}e{q}^{-1}$
• B. $67.95Sc{m}^{2}e{q}^{-1}$
• C. $543.6Sc{m}^{2}e{q}^{-1}$
• D. $135.9Sc{m}^{2}e{q}^{-1}$

Answer 1

Answer: (D)

${\lambda }_m^{\infty }={\lambda }_m^{\infty }$ (oxalate $)+{\lambda }_m^{\infty }\left(N{a}^{+}\right)+{\lambda }_m^{\infty }\left(K^{+}\right)$
$=(148\cdot 2+73\cdot 5+50\cdot 1)$
$=271.85c{m}^{2}mo{l}^{-1}$
$\because {\lambda }_{eq}^{\infty }=\frac{{\lambda }_m^{\infty }}{n\cdot \text{ factor }}=\frac{271.8}{2}$
$=135.9Sc{m}^{2}e{q}^{-1}$