To find the molar conductance of $CH_3COOH$ at infinite dilution, we will use Kohlrausch's Law of Independent Migration of Ions.$\\$According to Kohlrausch's Law, the molar conductance at infinite dilution for an electrolyte can be expressed as the sum of the molar conductances of its constituent ions.$\\$Given:$\\$• $\Lambda^\infty_{NaCl} = 126.45 \, \mathrm{S \, cm^2 \, mol^{-1}} \\$• $\Lambda^\infty_{HCl} = 426.16 \, \mathrm{S \, cm^2 \, mol^{-1}} \\$• $\Lambda^\infty_{CH_3COONa} = 91.0 \, \mathrm{S \, cm^2 \, mol^{-1}} \\$We need to find $\Lambda^\infty_{CH_3COOH}. \\$Using the law, we can write:$\\ \Lambda^\infty_{CH_3COOH} = \Lambda^\infty_{HCl} + \Lambda^\infty_{CH_3COONa} - \Lambda^\infty_{NaCl} \\$Substitute the given values:$\\ \Lambda^\infty_{CH_3COOH} = 426.16 + 91.0 - 126.45 \\$Calculate:$\\ \Lambda^\infty_{CH_3COOH} = 517.16 - 126.45 \\ \Lambda^\infty_{CH_3COOH} = 390.71 \, \mathrm{S \, cm^2 \, mol^{-1}} \\$Therefore, the molar conductance of $CH_3COOH$ at infinite dilution is $390.71 \, \mathrm{S \, cm^2 \, mol^{-1}}. \\$This corresponds to Option 2.