Question

Given that a 0.200 M HCl(aq) solution costs $39.95 for 500 mL, and that KCl costs $10/ton, which analysis procedure is more cost-effective?

Answer 1

1 L 1000 mL 200 mol HC 3.65 g x- -X- Amount of HCI = 500 mL x 1 L 1 mol HCl S 10.94 S/g 3.65 g Cost of HCl = 39.95 $ Cost of HCl1 Thus, 1 g HCl costs 10.94 S. Amount of KCl = 1 ton × 907 185 g Cost of KCI Thus, 1 g KCl costs 0.0000110 $/g Thus, analysis with KCl is cost effective 1 ton 100.0000110 S/g 907185g