Equivalent conductance of $NaCl, HCl$ and $C_{2} H_{5} COONa$ at infinite dilution are 126.45, 426.16 and 91 ${ohm}^{-} {cm}^{2}$
respectively. The equivalent conductance of $C_{2} H_{5} COOH$ is :

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$540.48 {ohm}^{-} {cm}^{2}$

$698.28 {ohm}^{-} {cm}^{2}$

$390.71 {ohm}^{-} {cm}^{2}$

$201 .28 {ohm}^{-} {cm}^{2}$

answer is B.

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Detailed Solution

We know that ;

$\begin{array}{l} \land_{C_{2} H_{5} COOH}^{\circ} = \land_{C_{2} H_{5} COONa}^{\circ} + \land_{HCl}^{\circ} - \land_{NaCl}^{\circ} \\ = (91 + 426 .16 - 126 .45) {Ohm}^{-} {cm}^{2} \\ = 390 .71 {ohm}^{-} {cm}^{2} \end{array}$

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