# How to interpret the instruction of the dilution of hydrogen peroxide?

I am required to prepare $$\pu{120cm3}$$ of hydrogen peroxide to each of one of my students tomorrow. Some material is given to me on the method of preparing the solutions. Within the method of preparing $$\ce{H2O2}$$, it says:

Prepare from freshly purchased $$100$$ volume ($$30\%~\text{w/w}$$, $$\pu{8.3 mol/dm3}$$) or $$20$$ volume ($$6\%~\text{w/w}$$, $$\pu{1.7 mol/dm3}$$) $$\ce{H2O2}$$. Dilute $$\pu{200 cm3}$$ of $$100$$ volume solution to $$\pu{1 dm3}$$ of $$20$$ volume solution, then dilute $$\pu{59 cm3}$$ of $$20$$ volume solution to $$\pu{1 dm3}$$ of $$\pu{0.1 mol/dm3}$$ solution.

Can someone please explain the quoted part?

Take $$\pu{200 mL}$$ of the $$30\%$$ solution, add water to a final volume of $$1$$ liter. Then you have a $$6\%$$ solution. Next, take $$\pu{59 mL}$$ of the $$6\%$$ solution and add water until it is $$1$$ liter.

After the first dilution it's not exactly $$6\%$$, because the density of $$30\%$$ $$\ce{H2O2}$$ is $$\pu{1.135 g/cm3}$$. It's more like $$6.6\%$$, but the instructions are referring to it as $$6\%$$. If you want exactly $$6\%$$ you should dilute by weight instead of volume.