Uestion 210.2 pts You have 5.00 mL of 2.26 M potassium bromide solution. If you add water until the total volume reaches 500.0 mL, what will be the molarity of potassium bromide in the resulting solution?

The answer to your question is Final concentration = 0.023 M

Explanation:

Data

Volume 1 = 5 ml

Concentration 1 = 2.26 M

Volume 2 = 500 ml

Concentration 2 = ?

Process

1.- To solve this problem use the dilution formula

   Concentration 1 x Volume 1 = Concentration 2 x Volume 2

- Solve for Concentration 2

     Concentration 2 = [Concentration 1 x Volume 1] / Volume 2

- Substitution

      Concentration 2 = [5 x 2.26] / 500

- Simplification

      Concentration 2 = 11.3 / 500

- Result

      Concentration 2 = 0.023 M

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

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