Chapters Included in SM: 3, 4, 5, 6, 9, 10, 11, 12, 13, 15, 16, 17, 18, 20, 21 and 23. Analytical Chemistry Hage/Carr 2 Solution Pen-Problem Chapter three (1) Change 1.04 g/mL to grams per Liter: 1.04 g/mL x 1000 mL/1L = 1004 grams/L. Then determine the grams of CH2O in the liter of solution: 1004 x 0.0552 = 55.42 g Next, change the gram/L to moles/L: 55.42 x 1 mol/30.03g = 1.846 M = 1.85 M (2) Determine the grams of water in the liter of the solution: 1004 – 55.42 = 948.6 g. Then convert grams of water to kg: 948.6 g x 1Kg/1000g = 0.9486 Kg. Divide moles in a liter by the kg of water in the liter: 1.85 moles/0.9486 Kg = 1.950 m