WBJEE · Chemistry · Electrochemistry
Equivalent conductivity at infinite dilution for \(\quad\) sodium-potassium oxalate \(\left[\left(\mathrm{COO}^{-}\right)_{2} \mathrm{Na}^{+} \mathrm{K}^{+}\right]\) will be [given molar
conductivities of oxalate, \(\mathrm{K}^{+}\) and \(\mathrm{Na}^{+}\) ions at infinite dilution are \(148.2, \quad 50.1,\) \(73.5 \mathrm{S} \mathrm{cm}^{2} \mathrm{mol}^{-1}\) respectively \(]\)
- A \(271.8 \mathrm{S} \mathrm{cm}^{2} \mathrm{eq}^{-1}\)
- B \(67.95 \mathrm{Scm}^{2} \mathrm{eq}^{-1}\)
- C \(543.6 \mathrm{Scm}^{2} \mathrm{eq}^{-1}\)
- D \(135.9 \mathrm{Scm}^{2} \mathrm{eq}^{-1}\)
Answer & Solution
Correct Answer
(D) \(135.9 \mathrm{Scm}^{2} \mathrm{eq}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\lambda_{m}^{-}=\lambda_{m}^{-}\) (oxalate) \(+\lambda_{m}^{-}\left(\mathrm{Na}^{+}\right)+\lambda_{m}^{-}\left(\mathrm{K}^{+}\right)\) \[ \begin{array}{l} =(148 \cdot 2+73 \cdot 5+50 \cdot 1) \\ =271.85 \mathrm{cm}^{2} \mathrm{mol}^{-1} \end{array} \] \(\therefore\)…
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