I am not sure if I calculated the exact molarity of the potassium permanganate. I multiplied the molarity of the potassium permanganate by the liters of potassium permanganate used in the titration and divided it by the total liters of solution. I don't know if that if how it is done or not.

(0.125M)(0.0378L)/0.0828L= 0.057M KMnO4

If you started with 37.8 mL of 0.125M KMnO4 and diluted it to 82.8 mL total volume, the new M is what you calculated. Without more details, however, I don't know if this will satisfy what you need to be doing.

To calculate the molarity of a solution using the formula you described, you need to make sure you are using the correct formula. The formula you provided is correct for calculating the final molarity of the potassium permanganate solution.

Here is the breakdown of the formula you used:

1. Multiply the molarity of the potassium permanganate (0.125 M) by the volume of potassium permanganate used in the titration (0.0378 L).
(0.125 M) x (0.0378 L) = 0.004725 moles of KMnO4

2. Divide this number by the total volume of the solution (0.0828 L).
0.004725 moles / 0.0828 L = 0.057 M KMnO4

So, based on your calculation, the molarity of the potassium permanganate solution is 0.057 M. Therefore, the formula and steps you used are correct.