The molar conductivities at infinite dilution | vedclass

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(D) According to Kohlrausch's law of independent migration of ions:$\Lambda_{m}^{\infty}(BaSO_{4}) = \lambda_{m}^{\infty}(Ba^{2+}) + \lambda_{m}^{\inf

Vedclass · Accepted Answer

$288$

Vedclass · Answer

(D) According to Kohlrausch's law of independent migration of ions:$\Lambda_{m}^{\infty}(BaSO_{4}) = \lambda_{m}^{\infty}(Ba^{2+}) + \lambda_{m}^{\infty}(SO_{4}^{2-})$We can express this in terms of the given electrolytes:$\Lambda_{m}^{\infty}(BaSO_{4}) = \Lambda_{m}^{\infty}(BaCl_{2}) + \Lambda_{m}^{\infty}(H_{2}SO_{4}) - 2\Lambda_{m}^{\infty}(HCl)$Substituting the given values:$\Lambda_{m}^{\infty}(BaSO_{4}) = 280 + 860 - 2(426)$$= 1140 - 852$$= 288 \, S \, cm^{2} \, mol^{-1}$

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