Anthony T. answered 03/21/21
271 ppm sulfate ion can be considered to be 271 g sulfate in 106 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 mg/mL / 96.1 mg / millimole x 1000 mL/L = 2.88 millimoles SO42- / Liter. I indicated which units cancel by a strike.
Always check math.