Question

The molar conductivity at infinite dilution of Al(SO4)3 is 858 S cm2/mol. Calculate the number of molar ionic conductivity of Al3+ ion given that lambda(SO4ion ) = 160 S cm2/mol.

Answer 1

Answer:

189 Scm^2 mol-1

I have not written the units in the image sorry

Answer 2

Answer:

Explanation:

Here in the question, we have been given that the molar conductivity at infinite dilution of Al(So₄)₃ is 858 cm²/mol and we are asked to find the molar ionic conductivity of Al³⁺ ion and given that λ(SO₄) ion = 160 cm²/mol.

Al₂(SO₄)  → 2Al³⁺ + 3SO₄²⁻

According to Kohlrausch's law:

                A⁰[AL₂(SO₄)₃] = 2λ⁰(AL³⁺)+3λ⁰(SO₄²⁻)

                      λ⁰(AL³⁺) = Λ⁰[AL₂(SO₄)₃]-3λ⁰(SO₄²⁻) / 2

                                    = $\frac{858-480}{2}$ = 189 cm² mol⁻¹

Hence the number of molar ionic conductivity of Al³⁺ ion is 189cm²mol⁻¹.