Question

The molar conductance at infinite dilution of Al₂(SO₄)₃ is 858 ohm⁻¹ cm² mol⁻¹. Calculate the molar conductance at infinite dilution of Al³⁺ ion if that of \[\ce{SO^{2-}_{4}}\] ion is 160 ohm⁻¹ cm² mol⁻¹.

Answer 1

Given: For the salt Al₂(SO₄)₃, it dissociates as:

\[\ce{Al2(SO4)3 −> 2Al^3+ + 3SO^{2−}_{4}}\]

Total molar conductance at infinite dilution:

\[\ce{\Lambda^0_m = 2 \lambda^0(Al^3+) + 3 \lambda^0(SO^{2−}_{4})}\]

\[\ce{\Lambda^0_m (Al2(SO4)3)}\] = 858 Ω⁻¹ cm² mol⁻¹

\[\ce{\lambda^0(SO^{2−}_{4})}\] = 160 Ω⁻¹ cm² mol⁻¹

858 = 2λ⁰ (Al³⁺) + 3 × 160

858 = 2λ⁰ (Al³⁺) + 480

858 − 480 = 2λ⁰ (Al³⁺)

378 = 2λ⁰ (Al³⁺)

λ⁰ (Al³⁺) = 189 Ω⁻¹ cm² mol⁻¹