The molar conductivities $\wedge_{\mathrm{NaOAC}}^{\circ}$ and $\wedge_{\mathrm{HCl}}^{\circ}$ at infinite…

The molar conductivities $\wedge_{\mathrm{NaOAC}}^{\circ}$ and $\wedge_{\mathrm{HCl}}^{\circ}$ at infinite dilution in water at $25^{\circ} \mathrm{C}$ are $91.0$ and $426.2 \mathrm{~S} \mathrm{~cm}^2 / \mathrm{mol}$ respectively. To calculate $\wedge_{\mathrm{HOAc}}^O$, the additional value required is
  1. $\wedge_{\mathrm{H}_2 \mathrm{O}}^{\circ}$
  2. $\wedge_{\mathrm{KCl}}^{\circ}$
  3. $\wedge_{\mathrm{NaOH}}^{\circ}$
  4. $\wedge_{\mathrm{NaCl}}^{\circ}$

Solution

$ \begin{aligned} & \lambda_{\mathrm{CH}_3 \mathrm{COONa}}^0=\lambda_{\mathrm{CH}_3 \mathrm{COO}^{-}}^0+\lambda_{\mathrm{Na}^{+}}^0 \\ & \lambda_{\mathrm{HCl}}^0=\lambda_{\mathrm{H}^{+}}^0+\lambda_{\mathrm{C}}^0 \cdots \cdots \cdots \cdots \cdot \\ & \lambda_{\mathrm{NaCl}}^0=\lambda_{\mathrm{Na}}^0+\lambda_{\mathrm{Cl}}^0 \cdots \cdots \cdots \cdots \\ & \lambda_{\mathrm{CH}_3 \mathrm{COOH}}^0=(1)+(2)-(3) \end{aligned} $

Asked in: JEE Main 2006

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