The limiting molar ionic conductivities of $Al X^{3 +}$ and $SO X_{4} X^{2 -}$ are $189$ and $160 S c m^{2} mo l^{- 1}$ respectively. The limiting equivalent conductivity of $Al X_{2} (SO X_{4}) X_{3}$ , in $S c m^{2} e q^{- 1}$ , is

Answer: A

By Kohlrausch's law,

Λ_{m}^{\circ} (Al X_{2} (SO X_{4}) X_{3}) = 2 λ^{\circ} (Al X^{3 +}) + 3 λ^{\circ} (SO X_{4} X^{2 -}) = 2 (189) + 3 (160) = 858 S c m^{2} mo l^{- 1}

. The n-factor of

Al X_{2} (SO X_{4}) X_{3}

is 6, so

Λ_{e q}^{\circ} = 858/6 = 143 S c m^{2} e q^{- 1}

. Adding the two ionic values once gives

349

, which ignores the stoichiometry.