For each solution below calculate the millimoles of solute.

a. 1.90 l of 0.00657 m nacn

b. 871 ml of a solution containing 8.49 ppm caco3

A

(1.62 L) x (0.00332 mol/L NaCN) = 0.00538 mol = 5.38 mmol NaCN

B

Supposing the solvent to be water, and supposing the density of the solution is near that of pure water:

(675 g) x (5.23 x 10^-6) / (100.0875 g CaCO3/mol) = 3.53 * 10^-5 mol = 0.0353 mmol CaCO3.