We are asked to calculate the molarity of the solution of the 56.2 mL BaCl2.
BaCl2 + Na2SO4 → BaSO4 + 2 NaCl
Molar mass Na2SO4 = 142.05 g/mol
If 56.2 mL of BaCl2 solution is needed to precipitate all the sulfate ion in a 752 mg sample of Na2SO4, what is the molarity of the solution?
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