A 0.4000 g sample that contained 96.4% Na2SO4 (142.04 g/mol) required 41.25 mL of barium chloride solution to reach the endpoint in a precipitation titration. What is the molarity of the BaCl2 solution?

0.0658 M

Explanation:

The reaction that takes place is:

Na₂SO₄(aq) + BaCl₂(aq) ↔ BaSO₄(s) + 2NaCl(aq)

First we calculate the number of Na₂SO₄ moles:

Now we convert to moles of BaCl₂:

Finally we divide by the volume to calculate the molarity:

the answer is true.

6.4 cm (max) - 25 cm (min)