Mn2+
ions is prepared by dissolving 1.584 g of pure manganese metal in nitric acid and diluting to a final volume of 1.000 L. The following solutions are prepared by dilution. For solution A, 50.00 mL of stock solution is diluted to 1000.0 mL. For solution B, 10.00 mL of A is diluted to 250.0 mL. For solution C, 10.00 mL of B is diluted to 500.0 mL. Calculate the molar concentrations of the stock solution and solutions A, B, and C.

The molarity of the stock Mn²⁺ ions is 0.0288 MBased on the dilution formula;The molarity of A is 0.00144 MThe molarity of B is 0.0000576 MThe molarity of C is 0.000001152 MWhat is the molarity of a solution?The molarity of a solution is the number of moles of a solute dissolved in a given volume of solution in liters.Molarity = number of moles/volumeThe molarity of the stock solution is:moles of Mn²⁺ ions = mass / molar massmolar mass of  Mn²⁺ ions = 55.0 g/molmoles of Mn²⁺ ions = 1.584 / 55moles of Mn²⁺ ions = 0.0288 molesmolarity of Mn²⁺ ions = 0.0288 / 1molarity of Mn²⁺ ions = 0.0288 MThe dilution formula is used to determine the molarities of A, B, and C.C₁V₁ = C₂V₂C₂ = C₁V₁ / V₂Where;C₁ = initial molarityV₁ = initial volumeC₂ = final molarityV₂ = final volumeMolarity of A = 50 * 0.0288 / 1000Molarity of A = 0.00144 MMolarity of B = 10 * 0.00144 / 250Molarity of B = 0.0000576 MMolarity of C = 10 * 0.0000576 / 500Molarity of C = 0.000001152 MLearn more about molarity at: https://brainly.com/question/17138838#SPJ1