How many mmol are in a 615mL solution containing 5.53 ppm CaCo3

Answer 1

part per million (ppm) is mg/L. so 5.53ppm is 5.53mg/L of CaCO3.

convert the concentration to Molarity (mol/L);
5.53mg = 5.53e-3g

mole in 5.53e-3g = m/Mr = 5.53e-3g/48gmol-1

= 1.152e-4mol

i.e. approximately 0.00012mol/L CaCO3.

to find the mole in 615mL, we multiply the molarity with the volume ensuring unit consistency in the volume;

i.e. 615mL = 0.615L

n = Mv = 0.00012molL-1x0.615L = 7.1e-5 mol.

mmol is a thousand mol therefore;

7.1e-5x1000 = 0.071mmol.

check the calculation carefully for any entry error.

hope that helps..