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 ]
A. A
Number of moles of Cu NO 3 2 = 3 . 74 187 = 0 . 02 2 Cu NO 3 2 + 4 KI → Cu 2 I 2 ↓ + I 2 + 4 KNO 3 Number of moles of Cu 2 l 2 precipitated = 0 . 01 I 2 brown   solution + H 2 S → S ↓ + 2 HI Number of moles of S precipitated = 0 . 01 Mass of S precipitates = 0 . 01 × 32 g = 0 . 32   g
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