JEE Advanced 2022 Paper 1 · Q04 · Qualitative Analysis
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]
Reveal answer + step-by-step solution
Correct answer:0.32
Solution
Molar mass of Cu(NO$_3$)$_2$ = 63 + 2(14 + 48) = 63 + 124 = 187 g mol$^{-1}$. Moles = 3.74/187 = 0.02 mol.
Reaction with excess KI: 2 Cu(NO$_3$)$_2$ + 4 KI $\rightarrow$ 2 CuI(s) + I$_2$ + 4 KNO$_3$. The I$_2$ produced dissolves in excess KI as KI$_3$ (the brown solution): I$_2$ + KI $\rightarrow$ KI$_3$. Net: 2 Cu(NO$_3$)$_2$ + 5 KI $\rightarrow$ 2 CuI$\downarrow$ + KI$_3$ + 4 KNO$_3$. From 0.02 mol Cu(NO$_3$)$_2$, KI$_3$ formed = 0.01 mol.
H$_2$S reduces the I$_3^-$ (i.e., the I$_2$ in solution) back to I$^-$, depositing sulphur: KI$_3$ + H$_2$S $\rightarrow$ S$\downarrow$ + 2 HI + KI. So X = S. Mass of X = 0.01 mol $\times$ 32 g mol$^{-1}$ = 0.32 g.
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