A stock solution containing Mn(2+) ions was prepared by dissolving 1.584 g pure manganese metal in nitric acid

Question

and diluting to a final volume of 1.000 L. wat is the concentration of the stock solution?

Answer 1

All you have to do is figure the number of moles of Mn in 1.584 g of Mn.

1.584 g Mn x (1 mol Mn / 54.938 g Mn) = 0.02883 mol Mn

Since the solution is diluted to 1 L, then the molarity is 0.02883 mol / L.