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Topic: Calculating weight needed to make a solution  (Read 2663 times)

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Offline GWashington1732

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Calculating weight needed to make a solution
« on: May 12, 2010, 01:02:21 PM »
I wanted to check my calculations before I turned this paper in.
I need to calculate the weight of KMnO4 required for 500 ml of 0.02 M solution.

So I found the number of grams per one mole.
(157.996 g/mole)
Multiplied that by the desired molarity
(157.996 g/mole) x (0.02 mole/L)
and multiplied by the amount of liters desired
so the equation ended as
(157.996 g/mole) x (0.02 mole/L) x (.5 L)= 1.57996 g

Offline Borek

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Re: Calculating weight needed to make a solution
« Reply #1 on: May 12, 2010, 01:36:49 PM »
Looks OK.

Watch number of significant figures.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline GWashington1732

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Re: Calculating weight needed to make a solution
« Reply #2 on: May 12, 2010, 01:40:59 PM »
Great, thanks.

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