The student diluted 25.0 cm3 of a solution of hydrogen peroxide with water and made the
solution up to 250.0 cm3. The student titrated 25.0 cm3 of this solution with 0.0200 mol dm–3
KMnO4 under acidic conditions. The volume of KMnO4(aq) required to reach the end-point was
23.45 cm3.
The overall equation for the reaction is given below.
2MnO4
– + 6H+ + 5H2O2 2Mn2+ + 8H2O + 5O2
• Calculate the concentration, in g dm–3, of the undiluted hydrogen peroxide solution.
I started by calculating the moles of KMnO4 23.45/1000 * 0.0200 = 4.69x10^-4
Then the moles of H2O2 in 25.0 by using the ratio 2:5 so moles of H2O2= 1.1725x10^-3 meaning there are 0.011725 moles in 250
How would you calculate the concentration in gdm-3? Don't understand this unit of concentraion?