In the Carius method for the estimation of halogens, 250 mg of an organic compound gave a certain mass of AgBr precipitate. If the percentage of bromine in the compound is 32 % , the mass of AgBr formed is _______ mg. (Given : molar mass in g mol ⁻¹ of Ag : 108, Br : 80 )

Step-by-step solution

Mass of bromine in the organic compound = 32 100 250 = 80 mg . Moles of bromine = 80 10⁻³ 80 = 10⁻³ mol . In the Carius method, 1 mole of Br produces 1 mole of AgBr . Moles of AgBr formed = 10⁻³ mol . Molar mass of AgBr = 108 + 80 = 188 g mol ⁻¹ . Mass of AgBr = 10⁻³ mol 188 g mol ⁻¹ = 0.188 g = 188 mg . Answer: 188