Answer to Question #126950 in Physical Chemistry for ALI

Question

4) Describe how you would prepare 1.0 L of a 0.10 M solution of sulfuric acid from a 3.0 M solution of sulfuric acid.
5) Describe how you would prepare 500 mL of a 1.0 M solution of potassium chloride that is 93.0% pure. The gram formula weight of potassium chloride is 74.56 g/mol.
6) Describe how you would prepare 1000 mL of a 1.0 N solution of magnesium hydroxide. The gram formula weight of magnesium hydroxide is 58.33 g/mo

Answer 1

4) C₁V₁ = C₂V₂ formula can be used to find the volume of the initial solution:

V₁ = C₂V₂ / C₁ = (1.0L x 0.10M) / 0.30M = 0.033 L

Therefore, 0.033 L or 33 mL of 3.0 M solution should be measured using the graduated cylinder, moved to the 1.0 L volumetric flask and diluted with water up to the 1.0 L mark.

5) 500 mL of 1.0 M KCl contains 1.0 x 0.5 = 0.5 moles of pure KCl.

m (KCl) = 0.5 mol x 74.56 g/mol = 37.28 g.

Considering the 93% purity of the sample, the actual mass weighted for preparation equals 37.28g / 0.93 = 40.09 g. Therefore, 40.09 g of the impure KCl should be weighted using the electronic balance, placed into the graduated glassware and water up to 500 mL mark should be added.

6) Assuming that Mg(OH)₂ is ideally soluble in water, the normality for this compound equals molarity x 2 because one mole of Mg(OH)₂ produces 2 moles of OH^- ions. Therefore the molarity of the solution is 1.0 / 2 = 0.5 M.

1000 mL of the solution contain 0.5 moles of Mg(OH)₂.

m (Mg(OH)₂) = 0.5 mol x 58.33 g/mol = 29.17 g.

So 29.17 g of Mg(OH)₂ should be weighted using the electronic balance, placed into the graduated glassware and water up to 1000 mL mark should be added.