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14 April, 12:33

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?

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  1. 14 April, 16:15
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    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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