0.2178 g of pure As2O3 (molar mass 198 g mol^-1) is dissolved, buffered with NaHCO3 and titrated with 0.100 M I2 (H3AsO3 is oxidised to H3AsO4). The volume of iodine solution needed is

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JEE Main levelSingle correct optionDifficulty 2 of 3eq083

$0.2178 g$ of pure $As X_{2} O X_{3}$ (molar mass $198 g mo l^{- 1}$ ) is dissolved, buffered with $NaHCO X_{3}$ and titrated with $0.100 M$ $I X_{2}$ ( $H X_{3} AsO X_{3}$ is oxidised to $H X_{3} AsO X_{4}$ ). The volume of iodine solution needed is

A22.0 mL
B44.0 mL
C11.0 mL
D5.5 mL

Answer: A

mol As2O3 = 0.2178/198 = 1.10e-03. Each As(III) loses 2 electrons, so As2O3 loses 4 e-, needing 2 mol I2 per mol. mol I2 = 2.20e-03; volume = 2.20e-03/0.100 = 0.0220 L = 22.0 mL.

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$0.2178 g$ of pure $As X_{2} O X_{3}$ (molar mass $198 g mo l^{- 1}$ ) is dissolved, buffered with $NaHCO X_{3}$ and titrated with $0.100 M$ $I X_{2}$ ( $H X_{3} AsO X_{3}$ is oxidised to $H X_{3} AsO X_{4}$ ). The volume of iodine solution needed is