# Chem

Potassium permanganate (KMnO4) solutions are used for the determination of iron in samples of unknown concentration. As a laboratory assistant, you are supposed to prepare 200. mL of a 0.300 M KMnO4 solution. What mass of KMnO4, in grams, do you need?

I got 60, but that's wrong. I know you have to get moles by using the molarity formula and then change it into grams using molar mass, but I can't get the right answer/ Any help is appreciated!

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1. You appear to know what to do; the problem may be that you're just making a math error. Post your work and I'll find the error for you.

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2. the molar mass of KMnO4 is

39.098 + 54.938 + 15.999*4 = 158.032

200 ml = 200 mL / 1000 mL/L = 0.2 L

0.2 L * 0.3 moles/L = 0.06 Moles KMnO4
= 0.06 * 158.032 g/mol = 9.48 g KMnO4

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3. You must have changed your method because you're no longer reporting 60. You're right, 60 is millimoles KMnO4 you need which is 0.06 mols. Then g = mols x molar mass = 0.06 x 158.034 = 9.48 g KMnO4.
That looks good to me.

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4. It says that the molarity is.3M so you would do .3 mol/1 L*.2L. .6mol*157g=94.2. That's my new answer.

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5. nope. 0.3 x 0.2 is right but that answer is 0.06 and not 0.6 and that x 158.34 (not 157) = 9.48g.

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