Which relation is correct for Lambda^0_{m(H_2 | vedclass

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(B) According to Kohlrausch's law of independent migration of ions, the molar conductivity of an electrolyte at infinite dilution is the sum of the mo

Vedclass · Accepted Answer

$\Lambda^0_{m(HCl)} + \Lambda^0_{m(NaOH)} - \Lambda^0_{m(NaCl)}$

Vedclass · Answer

(B) According to Kohlrausch's law of independent migration of ions, the molar conductivity of an electrolyte at infinite dilution is the sum of the molar conductivities of its constituent ions.For water $(H_2O)$, the dissociation is $H_2O \rightleftharpoons H^+ + OH^-$.Therefore, $\Lambda^0_{m(H_2O)} = \lambda^0_{H^+} + \lambda^0_{OH^-}$.We can obtain this by combining strong electrolytes:$\Lambda^0_{m(HCl)} = \lambda^0_{H^+} + \lambda^0_{Cl^-}$$\Lambda^0_{m(NaOH)} = \lambda^0_{Na^+} + \lambda^0_{OH^-}$$\Lambda^0_{m(NaCl)} = \lambda^0_{Na^+} + \lambda^0_{Cl^-}$By calculating $\Lambda^0_{m(HCl)} + \Lambda^0_{m(NaOH)} - \Lambda^0_{m(NaCl)}$, we get:$(\lambda^0_{H^+} + \lambda^0_{Cl^-}) + (\lambda^0_{Na^+} + \lambda^0_{OH^-}) - (\lambda^0_{Na^+} + \lambda^0_{Cl^-}) = \lambda^0_{H^+} + \lambda^0_{OH^-} = \Lambda^0_{m(H_2O)}$.Thus, option $(B)$ is correct.

Which relation is correct for $\Lambda^0_{m(H_2O)}$?

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