In Question 2, we calculated that it must have required 0.0005146 moles
of KMnO4 to react with the iron(II) contained in Sample 1. This
number of moles was contained in a volume of 25.11 mL (0.02511 L). So the
molarity (moles/L) of the potassium permanganate based on Sample 1 is given
by the following
Molarity of KMnO4 (Sample 1) = moles/L = 0.0005146 moles/0.02511
L = 0.02049 M
Since the procedure said that the KMnO4 solution would be
approximately 0.02 M when prepared correctly, this answer is consistent.