24.3cm^3 of 0.02moldm-3 KMnO4 reacted with 20cm^3 of iron (II) solution. Calculate the molarity of the iron (II) solution.

Firstly calcuate the moles of KMnO4 by using the equation n=MV/1000 where n=moles M=molarity (moldm^-3) and V= volume (cm³). As the question gives the molarity and volume of KMnO4 you simply plug the values into the equation to get n.

n of KMnO4 = (24.3 x 0.02)/1000 = 0.000486 = 4.86x10^-4

^{Now you know the moles of KMnO4 you can work out the number of moles of Fe2+ in solution.  (5Fe2+ + MnO4- + 8H+ -> 5Fe3+ + Mn3+ + 4H20) This equation is calculated from half equations. Therefore there is a 5:1 ratio of Fe2+ to MnO4-, so moles of Fe2+ = 0.000486 x 5= 0.00243}

^{You now have moles of Fe2+ and volume (20cm3) so you can rearrange the intial formula to get M=(1000 x n)/V therefore M=(1000 x 0.00243)/20 = 0.1215 moldm-3 as your final answer.}