We are asked to calculate the molarity of the solution of the 56.2 mL BaCl_{2}.

BaCl_{2 }+ Na_{2}SO_{4} → BaSO_{4} + 2 NaCl

Molar mass Na_{2}SO_{4} = 142.05 g/mol

If 56.2 mL of BaCl_{2} solution is needed to precipitate all the sulfate ion in a 752 mg sample of Na_{2}SO_{4}, what is the molarity of the solution?

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