Purple of Cassius is prepared by reducing AuCl₃ to colloidal gold by SnCl₂. A 1 L solution containing 1.97 mg of gold per ml is prepared from 0.05 M solution of AuCl₃ by reduction with appropriate amo

Question

Purple of Cassius is prepared by reducing AuCl₃ to colloidal gold by SnCl₂. A 1 L solution containing 1.97 mg of gold per ml is prepared from 0.05 M solution of AuCl₃ by reduction with appropriate amount of 0.05 M SnCl₂ solution, the resulting solution being diluted to 1 L with water. The volume of stannous chloride solution required, if its oxidation product is SnCl₄(aq), is (Au = 197)

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Mass of Au = 1.97 mg/ml x 1000 ml = 1.97 g = 1.97/197 = 0.01 mol. Au³⁺ + 3e⁻ → Au; Sn²⁺ → Sn⁴⁺ + 2e⁻, so mol SnCl₂ = 1.5 x 0.01 = 0.015 mol. Volume of 0.05 M SnCl₂ = 0.015/0.05 = 0.3 L = 300 ml, option (a).

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