Given Λ∞13Al3+=63Ω−1cm2mol−1 and λ∞12SO42−=80Ω−1cm2mol−1. The value Λ∞Al2SO43 would be

λ∞Al3+=3λ∞13Al3+=3×63Ω−1cm2mol−1=189Ω−1cm2mol−1λ∞SO42−=2λ∞12SO42−=2×80Ω−1cm2mol−1=160Ω−1cm2mol−1λ∞Al2SO43=2λ∞Al3++3λ∞SO42−=(2×189+3×160)Ω−1cm2mol−1=858Ω−1cm2mol−1

Given Λ∞13Al3+=63Ω−1cm2mol−1 and λ∞12SO42−=80Ω−1cm2mol−1. The value Λ∞Al2SO43 would be

λ∞Al3+=3λ∞13Al3+=3×63Ω−1cm2mol−1=189Ω−1cm2mol−1λ∞SO42−=2λ∞12SO42−=2×80Ω−1cm2mol−1=160Ω−1cm2mol−1λ∞Al2SO43=2λ∞Al3++3λ∞SO42−=(2×189+3×160)Ω−1cm2mol−1=858Ω−1cm2mol−1

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Given $Λ^{\infty} (\frac{1}{3} {Al}^{3 +}) = 63 Ω^{- 1} {cm}^{2} {mol}^{- 1}$ and $λ^{\infty} (\frac{1}{2} {SO}_{4}^{2 -}) = 80 Ω^{- 1} {cm}^{2} {mol}^{- 1}$ . The value $Λ^{\infty} ({Al}_{2} {({SO}_{4})}_{3})$ would be

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$143 Ω^{- 1} {cm}^{2} {mol}^{- 1}$

$858 Ω^{- 1} {cm}^{2} {mol}^{- 1}$

$206 Ω^{- 1} {cm}^{2} {mol}^{- 1}$

$286 Ω^{- 1} {cm}^{2} {mol}^{- 1}$

answer is B.

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Detailed Solution

$\begin{array}{l} λ^{\infty} ({Al}^{3 +}) = 3 λ^{\infty} (\frac{1}{3} {Al}^{3 +}) = 3 \times 63 Ω^{- 1} {cm}^{2} {mol}^{- 1} = 189 Ω^{- 1} {cm}^{2} {mol}^{- 1} \\ λ^{\infty} ({SO}_{4}^{2 -}) = 2 λ^{\infty} (\frac{1}{2} {SO}_{4}^{2 -}) = 2 \times 80 Ω^{- 1} {cm}^{2} {mol}^{- 1} = 160 Ω^{- 1} {cm}^{2} {mol}^{- 1} \\ λ^{\infty} ({Al}_{2} {({SO}_{4})}_{3}) = 2 λ^{\infty} ({Al}^{3 +}) + 3 λ^{\infty} ({SO}_{4}^{2 -}) = (2 \times 189 + 3 \times 160) Ω^{- 1} {cm}^{2} {mol}^{- 1} = 858 Ω^{- 1} {cm}^{2} {mol}^{- 1} \end{array}$