Question

Chloroquine (C18H26CIN3) has been widely promoted across the world as a drug that may treat the disease COVID-19. It is currently used as an anti-malarial drug and is soluble in dilute acids and slightly soluble in water. In the lab, a student wanted to dissolve a certain amount of this drug in a 0.01 mol/dm3 (M) hydrochloric acid (HCl) solution. The lab chemical stock room has a 37.0 % by mass of HCl stock solution, with a density of 1.2 g/cm3.

a) Calculate the molarity (M) of the 37.0 % by mass of HCl stock solution.

b) With the use of appropriate calculations/working, explain how the student would dilute the hydrochloric acid stock solution to a 250.0 cm3 of 0.01 M hydrochloric acid solution.

Answer 1

a)

100 g of this solution contains 37.0 g HCl

Volume of the solution: V = m/d = 100 / 1.2 = 83.33 mL

M (HCl) = 36.46 g/mol

n = 37.0 / 36.46 = 1.0148 mol

Molarity: C_M = n/V = 1.0148 / 0.08333 = 12.2 mol/L or 12.2 M

b)

250.0 mL 0.01 M solution contains: 0.01 * 0.250 = 0.0025 mol of HCl

If we use our stock solution we need:

(0.0025 mol * 1000 mL) / 12.2 M = 0.2 mL of the solution.

We withdraw this quantity with a special pipette into 250 mL volumetric flask and adjust the volume with distilled water.