Given
Molarity of Zn2+= 0.01M
Molarity of Cu2+ = 0.01M
Conc. of S2−=8.1∗10−21M
KspofZnS=3∗10−22
KspofCuS=8∗10−36
Solution
Ionic product of ZnS=[Zn2+]∗[S2−]
Ionic product of ZnS=0.01∗(8.1∗10−21)=8.1∗10−23
Ionic product of CuS=[Cu2+]∗[S2−]
Ionic product of CuS=0.01∗(8.1∗10−21)=8.1∗10−23
Difference of solubility product and ionic product of CuS is very large as compare to difference of solubility product and ionic product of ZnS.
Therefore CuS will precipitate.
The correct option is B