[GWashington1732]

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

[Borek]

Looks OK.

Watch number of significant figures.

[GWashington1732]

Great, thanks.