The treatment of an aqueous solution of 3.74 g of Cu(NO₃)₂ with excess KI results in a brown solution along with the formation of a precipitate. Passing H₂S 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]

The correct answer is 0.32.

Step 1: Calculate the molar masses of the relevant substances.
Given atomic masses:
Cu = 63, N = 14, O = 16, S = 32, I = 127, H = 1

Molar mass of Cu(NO₃)₂:
$\text{Molar mass of Cu(NO}$₂)₃=63+2×(14+3×16)=63+2×62=63+124=187 g/mol

Molar mass of Sulfur (S) = 32 g/mol

Step 2: Calculate the number of moles of Cu(NO₃)₂.

$\text{Moles of Cu(NO}$₃)₂=3.74 g187 g/mol=0.02 mol

Step 3: Analyze the chemical reactions taking place.
When aqueous Cu(NO₃)₂ is treated with excess KI, copper(II) ions react with iodide ions to form copper(I) iodide (Cu₂I₂ precipitate) and iodine (I₂), which imparts a brown color to the solution:

$2{\text{Cu}}^{2+}+4{\text{I}}^{-}\to {\text{Cu}}_2{\text{I}}_2\text{(precipitate)}+{\text{I}}_2\text{(brown solution)}$

From the stoichiometry of this reaction, 2 moles of Cu(NO₃)₂ (or Cu²⁺) produce 1 mole of I₂.
Therefore, 0.02 mol of Cu(NO₃)₂ will produce:

$\text{Moles of I}$₂=0.022=0.01 mol

Step 4: Reaction of the brown solution with H₂S.
When H₂S gas is passed through the brown solution containing iodine (I₂), hydrogen sulfide reduces iodine to iodide ions while getting oxidized to elemental sulfur (S), which forms precipitate X:

${\text{I}}_2+{\text{H}}_2\text{S}\to 2\text{HI}+\text{S}\downarrow \text{(precipitate X)}$

From the stoichiometry of this reaction, 1 mole of I₂ yields 1 mole of sulfur (S).
Therefore, 0.01 mol of I₂ produces 0.01 mol of sulfur (X).

Step 5: Calculate the mass of precipitate X (Sulfur).

$\text{Mass of X}=\text{Moles of S}\times \text{Molar mass of S}$

$\text{Mass of X}=0.01\text{mol}\times 32\text{g/mol}=0.32\text{g}$

Hence, the amount of precipitate X formed is 0.32 g.