In the laboratory, a student pipettes 25.0 mL of the same sodium sulfate solution into each of nine…

Step 1. Moles of precipitate. $n = \dfrac{1.165\ \text{g}}{233\ \text{g mol}^{-1}} = 0.00500\ \text{mol}$, that is $5.00\ \text{mmol}$.

Step 2. Sodium sulfate. Each mole of $\text{BaSO}_4$ needs one mole of $\text{Na}_2\text{SO}_4$, so the flask held $0.00500\ \text{mol}$ of it in $25.0\ \text{mL}$. Molarity $= \dfrac{0.00500\ \text{mol}}{0.0250\ \text{L}} = 0.200\ \text{mol L}^{-1}$.

Step 3. Barium chloride. At the break, $0.00500\ \text{mol}$ of $\text{BaCl}_2$ has been added in $20.0\ \text{mL}$. Molarity $= \dfrac{0.00500\ \text{mol}}{0.0200\ \text{L}} = 0.250\ \text{mol L}^{-1}$.

Step 4. Precipitate with $12.0\ \text{mL}$. This is before the break, so barium chloride limits. It brings $0.250\ \text{mol L}^{-1} \times 0.0120\ \text{L} = 3.00\ \text{mmol}$. Mass: $0.00300\ \text{mol} \times 233\ \text{g mol}^{-1} = 0.699\ \text{g}$.

Step 5. Chloride ions at the break. Each $\text{BaCl}_2$ gives $2$ chloride ions, which stay in solution: $2 \times 5.00 = 10.0\ \text{mmol}$. The volume is $25.0\ \text{mL} + 20.0\ \text{mL} = 45.0\ \text{mL}$. Molarity $= \dfrac{0.0100\ \text{mol}}{0.0450\ \text{L}} = 0.222\ \text{mol L}^{-1}$.