JEE Advanced 2022 — Paper 1 — CHEM-I solution

JEE Advanced2022Paper 1CHEM-I

The treatment of an aqueous solution of 3.74 g of Cu(NO $_{3}$ ) $_{2}$ with excess KI results in a brown solution along with the formation of a precipitate. Passing H $_{2}$ S through this brown solution gives another precipitate X. The amount of X (in g) is ________.

[Given: Atomic mass of H = 1, N = 14, O = 16, S = 32, K = 39, Cu = 63, I = 127]

Your answer

Solution

Molar mass of Cu(NO $_{3}$ ) $_{2}$ = 63 + 2(14 + 48) = 187 g/mol. Moles = $\frac{3.74}{187} = 0.02$ .

Step 1: Cu $^{2 +}$ oxidizes I $^{-}$ and precipitates as cuprous iodide while iodine is liberated:

$2 Cu(NO_{3})_{2} + 4 KI \to Cu_{2} I_{2} ↓ + I_{2} + 4 KNO_{3}$

From 0.02 mol Cu $^{2 +}$ , 0.01 mol I $_{2}$ is produced. Excess KI dissolves I $_{2}$ as the brown triiodide-type complex KI $_{3}$ .

Step 2: H $_{2}$ S reduces I $_{2}$ /I $_{3}^{-}$ , liberating elemental sulphur as the precipitate X:

$KI_{3} + H_{2} S \to KI + S ↓ + 2 HI$

Moles of S precipitated = moles of I $_{2}$ originally generated = 0.01.

Mass of S = $0.01 \times 32 = 0.32$ g.

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