Anthony T. answered • 03/21/21

Patient Math & Science Tutor

271 ppm sulfate ion can be considered to be 271 g sulfate in 10^{6} g of solution or 2.71 x 10^{-4} g sulfate for each gram of the solution.

As the solution density is 1.02 g/mL, the volume of one gram will be 1/1.02 = 0.980 mL/gm.

The weight-volume concentration of sulfate is then 2.71 x 10^{-4} g SO4 / g solution / 0.980 mL/g solution = 2.77 x 10^{-4} g SO4 / mL or 0.277 mg /mL.

The molecular mass expressed in mg is 96.1mg / millimole, so the millimolar concentration is

0.277 ~~m~~/__g__~~mL~~ / 96.1 ~~mg~~ / millimole x 1000 ~~mL~~/L = 2.88 millimoles SO4^{2-} / Liter. I indicated which units cancel by a strike.

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