Making a solid into a stock solution?

If a stock solution of Mn(2+) ions is prepared and the used to prepare other solutions by dilution, how do you?

• A stock solution containing Mn(2+) ions was prepared by dissolving 1.264g pure manganese metal in nitic acid and diluting to a final volume of 1.000L. The following solutions were then prepared by dilution. For solution A, 50.0mL of stock solution was diluted to 1000.0 mL. For solution B, 10.0 mL of solution A was diluted to 250.0mL. How do you calculate the concentration of the stock solution, solution A, and solution B? Please explain and show your work.

• Answer:

  (1.264 g Mn) / (54.9380 g Mn/mol) / (1.000 L) = 0.02301 mol/L Mn or Mn{2+} in the stock sol'n. (50.0 mL) × ( 0.02301 mol/L) / (1000.0 mL) = 0.00115 mol/L Mn or Mn{2+} in sol'n A. (10.0 mL) × (0.00115 mol/L) / (250.0 mL) = 4.60 × 10^-5 mol/L Mn or Mn{2+} in sol'n B.