1. Tardigrade
  2. Question
  3. Chemistry
  4. 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 ]

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The treatment of an aqueous solution of $3.74\, g$ of $Cu \left( NO _3\right)_2$ 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 $H =1, N =14, O =16, S =32, K =39, Cu =63, I =127$ ]

JEE AdvancedJEE Advanced 2022

Solution:

$ \underset{0.02}{2 Cu \left( NO _3\right)_2}+5 KI \longrightarrow Cu _2 I _2+ \underset{0.01}{KI _3}+4 KNO _3 $
$ \underset{0.01}{KI _3}+ H _2 S \longrightarrow \underset{0.01}{S \downarrow}+ KI +2 HI $
$ n _{ S }=0.01$ mole
weight of sulphur $=32 \times 0.01=0.32\, gm$