At 25^{circ} C,the molar conductances at infi | vedclass

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(B) According to Kohlrausch's law of independent migration of ions: $\lambda_{m, Ba(OH)_2}^{\infty} = \lambda_{m, Ba^{2+}}^{\infty} + 2\lambda_{m, OH^

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$524 	imes 10^{-4}$

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(B) According to Kohlrausch's law of independent migration of ions: $\lambda_{m, Ba(OH)_2}^{\infty} = \lambda_{m, Ba^{2+}}^{\infty} + 2\lambda_{m, OH^-}^{\infty}$ We can express this using the given electrolytes: $\lambda_{m, Ba(OH)_2}^{\infty} = \lambda_{m, BaCl_2}^{\infty} + 2\lambda_{m, NaOH}^{\infty} - 2\lambda_{m, NaCl}^{\infty}$ Substituting the given values: $\lambda_{m, Ba(OH)_2}^{\infty} = 280 	imes 10^{-4} + 2(248 	imes 10^{-4}) - 2(126 	imes 10^{-4})$ $= (280 + 496 - 252) 	imes 10^{-4}$ $= 524 	imes 10^{-4} \ S \ m^2 \ mol^{-1}$

At $25^{\circ} C$,the molar conductances at infinite dilution for the strong electrolytes $NaOH$,$NaCl$ and $BaCl_2$ are $248 \times 10^{-4}$,$126 \times 10^{-4}$ and $280 \times 10^{-4} \ S \ m^2 \ mol^{-1}$ respectively. The value of $\lambda_m^{\circ} Ba(OH)_2$ in $S \ m^2 \ mol^{-1}$ is:

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