0.800 g of a copper ore is dissolved and treated with excess KI (2Cu^2+ + 4I- -> 2CuI + I2). The iodine liberated needs 24.0 mL of 0.100 M Na2S2O3. Find the percentage of copper (Cu = 63.5) in the ore, to the nearest integer.

JEE Main levelNumerical answerDifficulty 2 of 3rx190

0.800 g of a copper ore is dissolved and treated with excess KI ( $2 Cu X^{2 +} + 4 I X^{-} 2 CuI + I X_{2}$ ). The iodine liberated needs 24.0 mL of 0.100 M $Na X_{2} S X_{2} O X_{3}$ . Find the percentage of copper (Cu = 63.5) in the ore, to the nearest integer.

Numerical answer type, answer in %, tolerance 0.5.

Each

Cu X^{2 +}

ends up equivalent to one thiosulphate (2 thiosulphate per

I X_{2}

, 2

Cu X^{2 +}

per

I X_{2}

). mmol Cu =

24.0 \times 0.100 = 2.40

; mass =

2.40 \times 63.5 = 152.4

mg. Percentage = 152.4/800 x 100 = 19.05%, about 19.