The molar conductivity at infinite dilution of barium chloride,...

From Kohlrausch's law

${Λ}_{m}^{\infty}\left({BaSO}_{4}\right) = {\lambda}_{m}^{\infty}\left({Ba}^{2 +}\right) + {\lambda}_{m}^{\infty}\left({SO}_{4}^{2 -}\right)$

${Λ}_{m}^{\infty}\left({BaSO}_{4}\right) = {Λ}_{m}^{\infty}\left({BaCl}_{2}\right) + {Λ}_{m}^{\infty}\left({H}_{2}{SO}_{4}\right) - 2{Λ}_{m}^{\infty} ( HCl )$

$= 280 + 860 - 2 ( 426 )$

$= 288{Scm}^{2}{mol}^{- 1}$