Use Cu = 63.5. A 0.500 g alloy sample is dissolved to give Cu^2+, treated with excess KI, and the liberated I2 (2Cu^2+ + 4I- -> 2CuI + I2) is titrated with 0.100 M Na2S2O3 (I2 + 2S2O3^2- -> 2I- + S4O6^2-); 18.0 mL is used. Find the mass percentage of Cu in the alloy (two decimals).

JEE Main levelNumerical answerDifficulty 2 of 3rx137

Use Cu = 63.5. A $0.500 g$ alloy sample is dissolved to give $Cu X^{2 +}$ , treated with excess $KI$ , and the liberated $I X_{2}$ ( $2 Cu X^{2 +} + 4 I X^{-} 2 CuI + I X_{2}$ ) is titrated with $0.100 M$ $Na X_{2} S X_{2} O X_{3}$ ( $I X_{2} + 2 S X_{2} O X_{3} X^{2 -} 2 I X^{-} + S X_{4} O X_{6} X^{2 -}$ ); $18.0 mL$ is used. Find the mass percentage of Cu in the alloy (two decimals).

Numerical answer type, answer in %, tolerance 0.05.

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

. Thiosulfate

= 18.0 \times 0.100 = 1.80

mmol = Cu. Mass

= 1.80 \times 1 0^{- 3} \times 63.5 = 0.1143 g

. Percentage

= 0.1143/0.500 \times 100 = 22.86%