Astudent has 2.99 g of potassium permaganate. how many ml of a .532 m solution can the student make?

how does one solve this?

Explanation:

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

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