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The ionic equivalent conductivity of $C_{2} O_{4}^{-2} {, K}^{+}$ and ${Na}^{+}$ ions are X, Y and Z ${S cm}^{2} {Eq}^{-1}$ respectively. The $\land_{e q}^{0}$ of $(NaOOC - COOK)$ based on the data given is

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$(x + \frac{z}{2} + y) {S Cm}^{2} {eq}^{-1}$

$(x + y+z) {S Cm}^{2} {eq}^{-1}$

$(x + \frac{y}{2} + z) {S Cm}^{2} {eq}^{-1}$

$(\frac{x}{2} + \frac{y}{2} + z) {S Cm}^{2} {eq}^{-1}$

answer is C.

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Detailed Solution

$\land^{o} eq$ of $(NaOOC - COOOK)$ = $\land_{e q}^{0} (o x a l a t e) + \land_{e q}^{0} (N a^{+}) + \land_{e q}^{0} (K^{+})$

= X+Y+Z S Cm²eq^-1

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