# The equivalent conductances of two strong electrolytes at infinite dilution in $\dpi{100} H_{2}O$( where ions move freely through a solution ) at 25°C are given below:$\dpi{100} \Lambda ^{\circ}{_{CH_{3}COONa}}= 91.0 S cm^{2}/equiv.$$\dpi{100} \Lambda ^{\circ}{_{HCl}}=426.2 S cm^{2}/equiv.$What additional information/quantity one needs to calculate $\dpi{100} \Lambda ^{\circ}$ of an aqueous solution of acetic acid? Option 1) $\dpi{100} \Lambda ^{\circ}$of chloroacetic acid $\left ( ClCH_{2}COOH \right )$ Option 2) $\dpi{100} \Lambda ^{\circ}$$of\: NaCl$ Option 3) $\dpi{100} \Lambda ^{\circ}$ $of \: CH_{3}COOK$ Option 4) The limiting equivalent conductance of  $H^{+}\left ( \lambda ^{\circ} {_{H^{+}}}\right )$

As we learnt in

Application of Kohlrausch's law -

Calculation of molar conductivities of weak electrolyte at infinite dilution.

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According to Kohlrausch’s law, the molar conductivity at infinite dilution $(\Lambda ^{\circ})$ for weak electrolyte, $CH_{3}COOH\; is$

$\Lambda ^{\circ}_{CH_{3}COOH}=\Lambda ^{\circ}_{CH_{3}COONa}+\Lambda ^{\circ}_{HCl}-\Lambda ^{\circ}_{NaCl}$

So, for calculating the value of  $\Lambda ^{\circ}_{CH_{3}COOH}$ , value  of $\Lambda ^{\circ}_{NaCl}$ should also be known.

Option 1)

$\dpi{100} \Lambda ^{\circ}$of chloroacetic acid $\left ( ClCH_{2}COOH \right )$

This option is incorrect.

Option 2)

$\dpi{100} \Lambda ^{\circ}$$of\: NaCl$

This option is correct.

Option 3)

$\dpi{100} \Lambda ^{\circ}$ $of \: CH_{3}COOK$

This option is incorrect.

Option 4)

The limiting equivalent conductance of  $H^{+}\left ( \lambda ^{\circ} {_{H^{+}}}\right )$

This option is incorrect.

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