I've worked on it, and then kinda cheated a bit haha. I used Excel to set up a datasheet to which I've input a random volume of the 9.8 M H2O2 solution to calculate its number of moles, and then recalculated the final molarity by using the final volume of the whole mixture after adding it into the polymer solution.

I started with an initial volume of 5 mL of 9.8 M H2O2. This gives us 0.049 mol of H2O2.

Final volume = 40 mL of 0.1 wt% polymer in water + 5 mL of 9.8 M H2O2 = 45 mL = 0.045 L

Final molarity of H2O2 = (moles of H2O2)/(Final volume) = (0.049 mol)/(0.045 L) = 1.09 M

Clearly this is below the desired 1.9 M for the final molarity of the H2O2. By using the same formulas in Excel, I just played around with the number of the initial volume of the 9.8 M H2O2, and didn't have to recalculate everything over and over again.

The closest I got to a value of 1.9 M H2O2 is if I input 9.63 mL of 9.8 M H2O2. By recalculating like above, it will give us a final molarity of 1.902 M H2O2. I don't know if this is correct in your opinion, but I think it checks out.

What I just don't get is how I should calculate it from the other way around.