Question

Equivalent conductivity at infinite dilution for sodium-potassium oxalate $\left(\right(CO{O}^{-}{)}_{2}N{a}^{+}{K}^{+})$ is :

Given : Molar conductivities of oxalate is $73.5Sc{m}^{2}mo{l}^{-1}$, $K^{+}$ and $N{a}^{+}$ ions at infinite dilution are $148.2,50.1Sc{m}^{2}mo{l}^{-1}$ 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

The correct option is D $135.9Sc{m}^{2}e{q}^{-1}$

${\stackrel{\infty }{\lambda }}_M={\stackrel{\infty }{\lambda }}_M\left(Oxalate\right)+{\stackrel{\infty }{\lambda }}_M\left(N{A}^{+}\right)+{\stackrel{\infty }{\lambda }}_M\left(K^{+}\right)$

${\stackrel{\infty }{\lambda }}_M=(148.+50.1+73.5)Sc{m}^{2}mo{l}^{-1}$

${\stackrel{\infty }{\lambda }}_M=271.8Sc{m}^{2}mo{l}^{-1}$

$\therefore {\stackrel{\infty }{\lambda }}_{Eq}=\frac{271.8}{2}=135.9Sc{m}^{2}e{q}^{-1}({\lambda }_{eq}^{\infty }=\frac{{\lambda }_M^{\infty }}{n.factor})$