2.50 g of crude CuSO4.5H2O (molar mass 249.5) is dissolved to make 250 mL of solution. A 25.0 mL aliquot is treated with excess KI (2Cu^2+ + 4I- -> 2CuI + I2) and the liberated iodine needs 9.00 mL of 0.100 M Na2S2O3. The percentage purity of the sample is:

JEE Advanced levelSingle correct optionDifficulty 3 of 3rx202

2.50 g of crude $CuSO X_{4} \cdot 5 H X_{2} O$ (molar mass 249.5) is dissolved to make 250 mL of solution. A 25.0 mL aliquot is treated with excess KI ( $2 Cu X^{2 +} + 4 I X^{-} 2 CuI + I X_{2}$ ) and the liberated iodine needs 9.00 mL of 0.100 M $Na X_{2} S X_{2} O X_{3}$ . The percentage purity of the sample is:

Thiosulphate =

9.00 \times 0.100 = 0.900

mmol = mmol

Cu X^{2 +}

in the aliquot (1:1). In 250 mL: 9.00 mmol. Mass =

9.00 \times 249.5 = 224.6

mg = 2.245 g. Purity = 2.245/2.50 x 100 = 89.8%.