The equivalent conductances of CH_3COONa,HCl, | vedclass

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(A) According to Kohlrausch's law of independent migration of ions,the equivalent conductance at infinite dilution for an electrolyte is the sum of th

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$126$

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(A) According to Kohlrausch's law of independent migration of ions,the equivalent conductance at infinite dilution for an electrolyte is the sum of the equivalent conductances of its constituent ions.For $CH_3COONa$: $\Lambda^\infty_{CH_3COONa} = \lambda^\infty_{CH_3COO^-} + \lambda^\infty_{Na^+} = 91 \ \Omega^{-1} \ cm^2 \ eq^{-1}$ $(i)$For $HCl$: $\Lambda^\infty_{HCl} = \lambda^\infty_{H^+} + \lambda^\infty_{Cl^-} = 426 \ \Omega^{-1} \ cm^2 \ eq^{-1}$ (ii)For $CH_3COOH$: $\Lambda^\infty_{CH_3COOH} = \lambda^\infty_{CH_3COO^-} + \lambda^\infty_{H^+} = 391 \ \Omega^{-1} \ cm^2 \ eq^{-1}$ (iii)We need to find $\Lambda^\infty_{NaCl} = \lambda^\infty_{Na^+} + \lambda^\infty_{Cl^-}$.By performing the operation $(i)$ + (ii) - (iii):$\Lambda^\infty_{NaCl} = (\lambda^\infty_{CH_3COO^-} + \lambda^\infty_{Na^+}) + (\lambda^\infty_{H^+} + \lambda^\infty_{Cl^-}) - (\lambda^\infty_{CH_3COO^-} + \lambda^\infty_{H^+})$$\Lambda^\infty_{NaCl} = \lambda^\infty_{Na^+} + \lambda^\infty_{Cl^-}$$\Lambda^\infty_{NaCl} = 91 + 426 - 391 = 126 \ \Omega^{-1} \ cm^2 \ eq^{-1}$.

The equivalent conductances of $CH_3COONa$,$HCl$,and $CH_3COOH$ at infinite dilution are $91$,$426$,and $391 \ \Omega^{-1} \ cm^2 \ eq^{-1}$ respectively. What is the equivalent conductance of $NaCl$ at infinite dilution?

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