Given: λ ∞ \(\left(\frac{1}{3} \mathrm{Al}^{3+}\right)\) = 63 Ω–1 cm2 mol–1 and λ ∞ \(\left(\frac{1}{2} \mathrm{SO}_4^{2-}\right)\) = 80 Ω–1 cm2 mol–1. The value of λ ∞ (Al2 (SO4)3) would be 

1
173 Ω–1 cm2 mol–1
2
858 Ω–1 cm2 mol–1
3
206 Ω–1 cm2 mol–1
4
286 Ω–1 cm2 mol–1
5
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