Question

What are the no of molecules of Iron iii in 285 g of FePO4​

Answer 1

Answer:

1.137 ×10^24 Fe3+

Explanation:

150.85 gm FePO4= 6.02×10^23 Fe3+

285 gm FePO4 = (6.02×10^23×285) /150.85 Fe3+

= 1.137 ×10^24 Fe3+