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?


1 Answer 1


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.


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