The  treatement  of an  Aqueous  solution of  3.74  gr of  $Cu{\left(N{O}_3\right)}_2$ 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  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)