If 56.0 ml of bacl2 solution is needed to precipitate all the sulfate ion in a 758 mg sample of na2so4, what is the molarity of the solution?

Answer is: molarity of solution is 0,0951 M.
Chemical reaction: BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl.
m(Na₂SO₄) = 758 mg ÷ 1000 mg/g = 0,758 g.
n(Na₂SO₄) = m(Na₂SO₄) ÷ M(Na₂SO₄).
n(Na₂SO₄) = 0,758 g ÷ 142 g/mol.
n(Na₂SO₄) = 0,00533 mol.
From chemical reaction: n(Na₂SO₄) : n(BaCl₂) = 1 : 1.
n(BaCl₂) = 0,00533 mol.
V(BaCl₂) = 56,0 mL = 0,056 L.
c(BaCl₂) = n(BaCl₂) ÷ V(BaCl₂).
c(BaCl₂) = 0,00533 mol ÷ 0,056 L.
c(BaCl₂) = 0,0951 mol/L.