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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: GWashington1732 on May 12, 2010, 01:02:21 PM

Title: Calculating weight needed to make a solution
Post by: GWashington1732 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

Title: Re: Calculating weight needed to make a solution
Post by: Borek on May 12, 2010, 01:36:49 PM

Looks OK.

Watch number of significant figures.

Title: Re: Calculating weight needed to make a solution
Post by: GWashington1732 on May 12, 2010, 01:40:59 PM

Great, thanks.