Question

Calculate the millimoles of solute:
901 mL of a solution containing 5.49 ppm CaCO3

Answer 1

5.49 ppm means 5.49 mg/L, So 901 ml of the solution contains:
5.49 mg/L ∙ 0.901 L = 4.94649 mg of CaCO3.
ν (CaCO3) = m / M = 4.94649 mg / 100 mg/mmol = 0.0494649 or approx. 0.05 mmol.