Part A) Approximately how many mL of 5% BaCl2 solution would berequired to precipitate all the sulfate if we assume that yoursamples are pure Magnesium sulfate? Assume that the density of thebarium chloride solution is 1.00 g/mL.

Part B) If the samples were pure potassium sulfate would yourequire a smaller or larger volume of barium chloride solution thanthe amount calculated in (Part A) above?

Crucible/sample Number	1	2	3
Mass of unknown sample (g)	0.3623 g	0.3461 g	0.3607 g
Mass of crucibles (g)	32.6072 g	33.0450 g	32.4736 g
Mass of crucible + BaSO4 (g)	32.8294 g	33.2525 g	32.6823 g
Mass of BaSO4 (g)	0.2222 g	0.2075 g	0.2087 g
percentage of SO42-	25.2 %	24.7 %	23.8 %

Average Percentage	24.6 %
                                                           St Dev	0.709

Mass of Sulfate,

Trial 1) 0.0914 grams

Trial 2) 0.0854 grams

Trial 3) 0.0859 grams

These were determined by stiochemetric calculations

0.2222g x (1 mole BaSO4 / 233.39 g) x (1 mole sulfate / 1 moeBaSO4) x (96.0576 g Sulfate / 1 mole sulfate) = 0.0914grams SO₄^2-

0.2075g x (1 mole BaSO4 / 233.39 g) x (1 mole sulfate / 1 moeBaSO4) x (96.0576 g Sulfate / 1 mole sulfate) = 0.0854grams SO₄^2-

0.2087g x (1 mole BaSO4 / 233.39 g) x (1 mole sulfate / 1 moeBaSO4) x (96.0576 g Sulfate / 1 mole sulfate)= 0.0859 gramsSO₄^2-

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Part A) Approximately how many mL of 5% BaCl2 solution would berequired to precipitate all the sulfate if we assume that yoursamples are pure Magnesium sulfate? Assume that the density of thebarium chloride solution is 1.00 g/mL.

Part B) If the samples were pure potassium sulfate would yourequire a smaller or larger volume of barium chloride solution thanthe amount calculated in (Part A) above?