A student has 2.99 g of Potassium Permaganate. How many mL of a. 532 M solution can the student make? How does one solve this?

you must be know the formula of potassium permaganate is KMnO₄

now you have already in question 2 things,

1. the molarity of solution = 0.532 M

2. the mass of KMnO₄ = 2.99 g

the molarity law is M = moles of solute / volume (L)

so use this equation to calculate the volume in liter and convert the result to milliliter.

(if the mass unit is gram (g) = volume must be litter (L) and if the mass unit is milligram = volume must be milliliter (ml))

* * first you must use this equation to calculate how many moles do you have in KMnO₄

number of moles = mass (g) / molecular weight "MW" (g/mole)

now you have 2.99 g (Mentioned in Question)

and you can calculate MW if you now the Chemical formula

so we have KMnO₄

Mw for KMnO₄ = 39 + 55 + 4x16 = 158 g/mole

* * (K=39, Mn=55, O = 16)

so number of moles = 2.99 / 158 = 0.018 = 0.02 moles

and if you go back to the molarity law:

Molarity = moles of solute / volume (L)

so 0.532 = 0.02 / volume (L)

Volume (L) = 0.532 x 0.02 = 0.01064 L

convert to ml = 0.01064 x 1000 = 10.64 ml

Good Luck