Question

Calculate the final molarity of H2O2 if 1.3 mL of the 3.0% w/w H2O2 solution is diluted with 3.7 mL water, then added to 5.0 mL of a starch-iodide solution.

Answer 1

Initial concentration = 3% w/w = 3 g of H₂O₂/100 g of solution

Molar mass of hydrogen peroxide = 34.02 g/mol

Density of 3% H₂O₂ = 1 g/ml

Volume of solution = mass of solution/density = 100 g/1 g/ml = 100 ml = 0.1 l (1000 ml = 1 lt)

No. of moles of H₂O₂ = mass/molar mass = 3 g/34.02 g/mol = 0.0882 moles

Initial molarity, M1 = no. of moles of H₂O₂/Vol in lts = 0.0882 moles/0.1 lt = 0.882 M

Initial volume, V1 = 1.3 ml

Final volume, V2 = 1.3 ml H₂O₂ + 3.7 ml H₂O+ 5 ml starch iodide solution = 10 ml

Final molarity, M2 = M

M1V1 = M2V2

0.882 M x 1.3 ml = M x 10 ml

1.1466/10 = M

Final molarity, M = 0.11466 = 0.115 M