To determine the molarity of the BaCl2 solution, we first need to understand the reactions involved. Barium chloride (BaCl2) reacts with sodium sulfate (Na2SO4) to produce barium sulfate (BaSO4), a precipitate. The balanced chemical equation for this reaction is:
\[ \text{Ba}^{2+} (aq) + \text{SO}_4^{2-} (aq) \rightarrow \text{BaSO}_4 (s) \]
Step 1: Calculate moles of Na2SO4.
Given:
- Mass of Na2SO4 = 758 mg = 0.758 g.
Molar mass of Na2SO4:
\[ \text{Molar mass of Na} = 22.99 , \text{g/mol}, \quad \text{Molar mass of S} = 32.07 , \text{g/mol}, \quad \text{Molar mass of O} = 16.00 , \text{g/mol} \]
\[ \text{Molar mass of Na2SO4} = (2 \times 22.99) + 32.07 + (4 \times 16.00) \ = 45.98 + 32.07 + 64.00 = 142.05 , \text{g/mol} \]
Calculate moles of Na2SO4:
\[ \text{Moles of Na2SO4} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.758 , \text{g}}{142.05 , \text{g/mol}} \approx 0.00534 , \text{mol} \]
Step 2: Moles of BaCl2 required.
From the balanced equation, we see that the moles of BaCl2 required is equal to the moles of Na2SO4, since 1 mole of Na2SO4 reacts with 1 mole of BaCl2:
\[ \text{Moles of BaCl2} = \text{Moles of Na2SO4} = 0.00534 , \text{mol} \]
Step 3: Calculate the molarity of BaCl2 solution.
Volume of BaCl2 solution = 57.0 mL = \(0.0570, \text{L}\).
Molarity (M) is defined as:
\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
\[ \text{Molarity of BaCl2} = \frac{0.00534 , \text{mol}}{0.0570 , \text{L}} \approx 0.09368 , \text{M} \]
Step 4: Express the molarity to three significant digits.
\[ \text{Molarity of BaCl2} \approx 0.0937 , \text{M} \]
Thus, the molarity of the BaCl2 solution is 0.0937 M.