Use Cu = 63.5. A copper(II) solution is treated with excess KI (2Cu^2+ + 4I- -> 2CuI + I2) and the iodine is titrated with 0.1 M Na2S2O3 (I2 + 2S2O3^2- -> 2I- + S4O6^2-). If 20 mL of thiosulfate is used, the mass of copper in the sample is:

JEE Main levelSingle correct optionDifficulty 2 of 3rx126

Use Cu = 63.5. A copper(II) solution is treated with excess $KI$ ( $2 Cu X^{2 +} + 4 I X^{-} 2 CuI + I X_{2}$ ) and the iodine is titrated with $0.1 M$ $Na X_{2} S X_{2} O X_{3}$ ( $I X_{2} + 2 S X_{2} O X_{3} X^{2 -} 2 I X^{-} + S X_{4} O X_{6} X^{2 -}$ ). If $20 mL$ of thiosulfate is used, the mass of copper in the sample is:

2 Cu X^{2 +} \to 1 I X_{2} \to 2 S X_{2} O X_{3} X^{2 -}

, so

Cu X^{2 +} : S X_{2} O X_{3} X^{2 -} = 1 : 1

. Moles of thiosulfate

= 20 \times 0.1 = 2

mmol, so Cu

= 2

mmol

= 2 \times 1 0^{- 3} \times 63.5 = 0.127 g