The treatement of an Aqueous solution of 3.74 gr of $C u {(N O_{3})}_{2}$ 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 number of Atoms of ‘x’ precipitate is $6.023 \times 10^{y}$ Then $\frac{y}{3}$ is (At. Wt of Cu=63,S=32, N=14 , O=16 , H=1)

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 7.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Question Image
No. of moles of $C u {(N O_{3})}_{2}$ = $\frac{3.74}{187} = 0.02$

2 moles of $C u {(N O_{3})}_{2}$ gives 1 mole of ‘S’ ppt
0.02 moles of $C u {(N O_{3})}_{2}$ gives 0.01mole of ‘S’ ppt= $6.023 \times 10^{21}$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE