Question

2. Calculate the molarity of a solution made from 15.00 mL of 32.0 wt. % HCl stock (density of 1.161 g/mL) diluted to 100 mL in a volumetric flask. Calculate the spike volume of that 100 mL dilute solution needed to make 1.000 L of 1.25 x 10 ppm HCl solution.

Answer 1

(page- 2. . Mass of 15.00 mL stock solution = 15.00 mL x 1.161 - 17.415 g Now, 100 g solution contains ::17.415g 32.0 g HCI. 32.0 x 17.415 g HU 100 gm - 5.5728 g HC Ablen dilution, 100 mL diluted solution has 5.5728 g HCI .: 1000 u 5.5728X1000 a HCA g Her 100 = 55,728 g HCI - 55.728 g HCL 36.45 g/mol = 1.529 mol HCI : Molarity of the solution = 1.529 M

du page-2 ppm = 1 mg/l :: 1.000 L ob 1.25 Ⓡ102 ppm HCl solution contains 1.25 x102 cl. Now, 5.5728 x103 mg HC is present in 100 mL dilute solution. 1.25 x102 mg HCl is present in tin 100x 1.25X102 mL dilecte solution. 5.5728 x103 = 2.243 mL dilute solution :: The answer is 2.243 mL).