A quantity of 9.8 g of FeSO₄·(NH₄)₂SO₄·xH₂O was dissolved in 250 ml of its solution. Twenty millilitres of this solution requires 20 ml of KMnO₄ solution containing 3.53 g of 90% pure KMnO₄ dissolved

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A quantity of 9.8 g of FeSO₄·(NH₄)₂SO₄·xH₂O was dissolved in 250 ml of its solution. Twenty millilitres of this solution requires 20 ml of KMnO₄ solution containing 3.53 g of 90% pure KMnO₄ dissolved per litre. The value of 'x' is

TestHub · Accepted Answer

Pure KMnO₄ per litre = 3.53 x 0.90 = 3.177 g → 3.177/158 = 0.0201 M. mmol KMnO₄ in 20 ml = 0.402; in acid, MnO₄⁻:Fe²⁺ = 1:5, so Fe²⁺ in the 20 ml aliquot = 2.01 mmol, i.e. 25.1 mmol in 250 ml. Molar mass of the hydrate = 9.8/0.0251 = 390. Anhydrous FeSO₄·(NH₄)₂SO₄ = 284, so 18x = 390 - 284 = 106 ≈ 108, x = 6, option (c).

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