In: Chemistry
a sample aqueous solution of iron(iii) sulfate has a mass of 200g, a density of 1.05 g/cm^3 and 8.35 x10^21 atoms of sulfur. What is the molarity of the solution?
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*10^23 = 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*10^23 = 1.85g of Fe2(SO4)3
volume of solution = mass of solution /density
= 200g/1.05g/cm^3 = 190.5cm^3
molarity = W*1000/G.M.Wt* volume of solution in ml
= 1.85*1000/400*190.5 = 0.0243M >>>>answer