Fe2(SO4)3
1mole of Fe2(SO4)3 contains 3 moles of S
6.023*10^23 molecules of Fe2(SO4)3 contains 3* 6.023*10^23 of
S
3 * 6.023*10^23 molecules of S present in 6.023*10^23 molecules
of Fe2(SO4)3
8.35*10^21 molecules of S present in
6.023*10^23*8.35*10^21/3*6.023*10^23Â Â Â = 2.78*10^21
molecules of Fe2(So4)3
6.023*10^23 molecules of Fe2(SO4)3 = 400g
2.78*10^21 molecules of Fe2(SO4)3 =
400*2.78*10^21/6.023*10^23Â Â = 1.85g of Fe2(SO4)3
volume of solution = mass of solution /density
                            Â
= 200g/1.05g/cm^3Â Â Â = 190.5cm^3
molarity =Â Â W*1000/G.M.Wt* volume of solution in
ml
              Â
= 1.85*1000/400*190.5 = 0.0243M >>>>answer
