The treatment of an aqueous solution of 3.74 g of \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) with excess KI results in a brown solution along with the formation of a precipitate. Passing \(\mathrm{H}_{2} \mathrm{~S}\) through this brown solution gives another precipitate X. The amount of X (in g) is ____. [Given : Atomic mass of \(\mathrm{H}=1, \mathrm{~N}=14, \mathrm{O}=16, \mathrm{~S}=32, \mathrm{~K}=39, \mathrm{Cu}=63, \mathrm{I}=127\) ]

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Practical and Analytical Chemistry

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

]

Answer: 0.32

Moles of

Cu (NO_{3})_{2} = \frac{3.74}{188} = 0.02

mol.

The reaction with KI:

2 Cu (NO_{3})_{2} + 5 KI \to Cu_{2} I_{2} + KI_{3} + 4 KNO_{3}

From 0.02 mol

Cu (NO_{3})_{2}

, we get 0.01 mol

KI_{3}

(brown solution).

Passing

H_{2} S

through

KI_{3}

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

From 0.01 mol

KI_{3}

, we get 0.01 mol S.

Mass of S =

0.01 \times 32 = 0.32

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Numerical answer