The steps required to prepare 200.0 ml of an aqueous solution of iron (iii) chloride, at a concentration of 1.25x10^-2 m. show the calculations determining what mass of solute to use in making this solution.

In order to prepare 200.0 mL of an aqueous solution of iron (III) chloride, at a concentration of 1.25 x 10⁻² M, you need to weight 0.4055 g of FeCl₃ and add to 200.0 mL of water.

Explanation:

Concentration: 1.25 x 10⁻² M

1,25 x 10⁻² mol FeCl₃ ___ 1000 mL

              x                   ___ 200.0 mL

         x = 2.5 x 10⁻³ mol FeCl₃

Mass of FeCl₃:

1 mol FeCl₃ 162.2 g

2.5 x 10⁻³ mol FeCl₃    y

                  y = 0.4055 g FeCl₃

Sodium phosphate has a chemical symbol of Na₃PO₄. When this dissociates into its ionic components then, these are Na⁺ and PO₄³⁻. The first thing that needs to be done is to divide the given mass of the substance by its molar mass to reveal how many moles of substance is present as shown below.

 molar mass of Na₃PO₄ is 163.94 g/mol

        number of moles = (20 g)(1 mol / 163.94 g) = 0.122 moles

Then, multiply the number of moles by the Avogadro's number to reveal how many formula units are present.

         number of formula units = (0.122 moles)(6.022 x 10²³ formula units/mol)
       number of formula units = 7.35 x 10²²

Since, in every formula unit there are 3 Na⁺ then, the answer should be equal to,
        number of Na⁺ ions = (3)(7.35 x 10²²) = 2.2 x 10²³ ions

2.2 x 10²³ ions