Question

4. A student makes the following preparation. She takes 400 grams of chemical A and add water to a final volume of 1 Liter. The molecular weight of chemical A is 200 grams/mole. This solution is referred to as solution 1. She then takes 2 ml of solution 1 and adds it to 8 mls of water and mixes it; this is solution 2. She then take 5 ml of solution 2 and adds it to 5 mls of a water and mixes it; this is solution 3. She then takes 1 ml of solution 3 and adds it to 9 mls of water and mixes it; this is solution 4. What is the concentration of A in solution 4? Note units. A) 1 mM B) 4 mm C) 10 mm D) 20 mM E) 40 mM

**Answer D: Please show steps and explanations for each step**

Answer 1

(4) Concentration of solution 1 = Mass Molar mass x volume (14) 400g - 200 g/mol X IL = 2 mol/L = 2M Concentration of solution 1 : 2M Solution 1 will be the stock dilute solution from stock relation is used . solution. solution To make following Mstock Vstock = Mdilute Vdilute/ Where: Mstock = Concentration of Stock solution. Ustock & volume of stock solution needed - to prepare dilute solution Vidilute & volume of dilute solution Mailute - Concentration of dilute solution. of Preparation Volume solution 2 : dilute solution of : 2 mL + 8 mL = loml Mstock Vstock = Mdilute Vailute 2M X 2 mL = Mailute x 10mL Mdilute = 2 x 2 mL 10mL Mailute = 0.4m

Preparation of solution (3) Concentration of solution 2 will be considered as concentration of stock solution for preparation of 3. Volume of stock solution = 5 mL Mstock = 0.4M Vailute = 5mL +5mL = 10mL (stock) (H₂O) 0.4m x 5mL = Mailute & lomL. # I Mailute = 0.4m x 5 mL - 0. 2M.. 10mL Preparation of solution 4: Mstock - 0.2M Vstock - 1 mL. Vailute = ImL +9mL = 10mL. 0.2 M x 1 mL = Mailute & loml Mailut - 0.2 M x ImL 10mL = 0.02 M. Concentration of solution 4 : 0.02m. IM = 10+3 mm. 0.02M = 10+3 mm x 0.02 M 1M Concentration of solution 4.20mm ).