Ok, so here's my question:

If exactly 14.563 grams of KMnO4 were dissolved to yield 750 mL of solution waht would be the normality of the KMnO4? From a previous question (that this question says to refer to) we have:

MnO4- +8H+ +5e --> Mn2+ +4H2O

I'm not exactly sure what to do. I know that N=#eq/L but that's it...Any help would be lovely.

Well, I see 5 equivalents per molecule of KMnO4

Normality= 5*molarity

calculate molarity and you have it.

You have to be careful on potassium permanganate, in caculating equivalents, because the oxidation change for Mn is usually different in each reaction.

Ok, I see.

To find the normality of KMnO4, you need to determine the number of equivalents (eq) of KMnO4 in the given amount of the compound. Here's how you can calculate it:

1. Start by finding the number of moles of KMnO4 using its molar mass. The molar mass of KMnO4 is:
M(KMnO4) = (39.10 g/mol) + (54.94 g/mol) + (4 * 16.00 g/mol) = 158.04 g/mol.

Number of moles of KMnO4 = mass / molar mass = 14.563 g / 158.04 g/mol = 0.09212 mol.

2. Next, you need to determine the number of equivalents of KMnO4. The reaction provided indicates that one KMnO4 molecule donates five electrons.

Number of equivalents of KMnO4 = number of moles of KMnO4 × number of electrons transferred = 0.09212 mol × 5 eq/mol = 0.4606 eq.

3. Finally, calculate the normality of KMnO4 by dividing the number of equivalents by the volume in liters.

Normality (N) = number of equivalents / volume (in L) = 0.4606 eq / 0.75 L = 0.614 N.

Therefore, the normality of the KMnO4 solution is 0.614 N.