Question:
To determine the sulfate concentration in a liquid fertilizer (aqueous solution), an analyst first diluted 50 mL of the fertilizer solution to 250 mL, pipetted out 25 mL of the dilute solution, added 200 mL water to it then precipitated the sulfate with barium nitrate solution. The precipitate was filtered, washed and dried to constant mass. In repeated experiments the mass of precipitate was 0.728 g, 0.773 g, 0.722 g, 0.732 g. As accurately as you can, calculate the percentage sulfate in the original fertilizer solution. Justify the procedure you followed.


what I did: By ignoring the second value, the average mass of the fertiliser is 0.727g.
moles of sulfate in barium sulfate = moles of barium sulfate
so n(SO4-) = 0.727/233.37 = 0.00312 moles in the dilute solution.

the additional 200ml doesn't affect this. if 25 ml of the dilute solution contains 0.00312 moles of barium sulfate, then 250mL would contain 10 times as much; 0.0312moles. all this comes from 50ml of the fertiliser solution; there are 0.0312moles of barium sulfate in the original fertiliser solution.

mass of sulfate in original fertiliser solution = m x MM = 0.0312 x (32.07 +4 x 16) = 2.99g
So, % = 2.99g/0.05L x 100 = 5985.59% ?? the answer is 5.99% (w/v). did they leave the units for 50mL as mL? shouldn't you convert to SI units, which is litres, not millilitres?