The treatment of an aqueous solution of $3.74 \mathrm{g}$ of $\mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2$ with excess $\mathrm{KI}$ results in a brown solution along with the formation of a precipitate. Passing ${\mathrm{H}}_2\mathrm{S}$ through this brown solution gives another precipitate $\mathrm{X}$. The amount of $\mathrm{X}$ (in $\mathrm{g}$) is____[Given: Atomic mass of $\mathrm{H}=1, \mathrm{N}=14, \mathrm{O}=16, \mathrm{S}=32, \mathrm{K}=39, \mathrm{Cu}=63, \mathrm{I}=127$]

Hence, the correct answer is 0.32.