25.0 mL of 0.100 M CuSO4 is treated with excess KI (2Cu^2+ + 4I^- -> 2CuI + I2). Taking I = 127 and Cu = 63.5, which of the following are correct?

1.25x10^-3 mol of I2 is liberated.

25.0 mL of 0.100 M Na2S2O3 is needed to titrate the iodine.

5.0x10^-3 mol of KI is consumed in total.

The mass of CuI precipitated is 0.953 g.

1.25x10^-3 mol of I2 is liberated.

25.0 mL of 0.100 M Na2S2O3 is needed to titrate the iodine.

5.0x10^-3 mol of KI is consumed in total.

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JEE Advanced levelOne or more correct optionsDifficulty 3 of 3eq090

$25.0 mL$ of $0.100 M$ $CuSO X_{4}$ is treated with excess $KI$ ( $2 Cu X^{2 +} + 4 I X^{-} 2 CuI + I X_{2}$ ). Taking I = 127 and Cu = 63.5, which of the following are correct?

AThe mass of $CuI$ precipitated is $0.953 g$ .
B $1.25 \times 1 0^{- 3}$ mol of $I X_{2}$ is liberated.
C $25.0 mL$ of $0.100 M$ $Na X_{2} S X_{2} O X_{3}$ is needed to titrate the iodine.
D $5.0 \times 1 0^{- 3}$ mol of $KI$ is consumed in total.

Answer: B, C, D

mol Cu2+ = 2.5e-3. I2 = 1.25e-3 mol; thiosulphate = 2.5e-3 mol = 25.0 mL of 0.100 M. KI consumed = 2 mol per mol Cu = 5.0e-3 mol. CuI = 2.5e-3 x 190.5 = 0.476 g, not 0.953 g.

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25.0 mL of 0.100 M CuSOX4​ is treated with excess KI (2CuX2++4IX−​2CuI+IX2​). Taking I = 127 and Cu = 63.5, which of the following are correct?

Answer: B, C, D

More questions on Iodometric and iodimetric titrations

$25.0 mL$ of $0.100 M$ $CuSO X_{4}$ is treated with excess $KI$ ( $2 Cu X^{2 +} + 4 I X^{-} 2 CuI + I X_{2}$ ). Taking I = 127 and Cu = 63.5, which of the following are correct?