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

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(B) According to Kohlrausch's law of independent migration of ions,the molar conductivity at infinite dilution for $Ba(OH)_2$ can be expressed as: $\L

Vedclass · Accepted Answer

$524 	imes 10^{-4}$

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(B) According to Kohlrausch's law of independent migration of ions,the molar conductivity at infinite dilution for $Ba(OH)_2$ can be expressed as: $\Lambda_{m, Ba(OH)_2}^{\infty} = \Lambda_{m, Ba^{2+}}^{\infty} + 2\Lambda_{m, OH^-}^{\infty}$ Given values: $\Lambda_{m, KOH}^{\infty} = \Lambda_{m, K^+}^{\infty} + \Lambda_{m, OH^-}^{\infty} = 248 	imes 10^{-4}\,S\,m^2\,mol^{-1}$ $\Lambda_{m, KCl}^{\infty} = \Lambda_{m, K^+}^{\infty} + \Lambda_{m, Cl^-}^{\infty} = 126 	imes 10^{-4}\,S\,m^2\,mol^{-1}$ $\Lambda_{m, BaCl_2}^{\infty} = \Lambda_{m, Ba^{2+}}^{\infty} + 2\Lambda_{m, Cl^-}^{\infty} = 280 	imes 10^{-4}\,S\,m^2\,mol^{-1}$ Using the relation: $\Lambda_{m, Ba(OH)_2}^{\infty} = \Lambda_{m, BaCl_2}^{\infty} + 2\Lambda_{m, KOH}^{\infty} - 2\Lambda_{m, KCl}^{\infty}$ Substituting the values: $= 280 	imes 10^{-4} + 2(248 	imes 10^{-4}) - 2(126 	imes 10^{-4})$ $= (280 + 496 - 252) 	imes 10^{-4}$ $= 524 	imes 10^{-4}\,S\,m^2\,mol^{-1}$

At $25\,^{\circ}C$,the molar conductivity at infinite dilution for electrolytes $KOH$,$KCl$ and $BaCl_2$ are $248 \times 10^{-4}$,$126 \times 10^{-4}$ and $280 \times 10^{-4}\,S\,m^2\,mol^{-1}$ respectively. Calculate $\Lambda _m^{\infty}$ for $Ba(OH)_2$ (in $S\,m^2\,mol^{-1}$).

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