What volume of 5 M Na₂SO₄ must be added to 25 mL of 1 M BaCl₂ to produce 10 g of BaSO₄ ?

Na₂SO₄ + BaCl₂ → BaSO₄ + 2NaCl
No. of moles of BaSO₄ $=\frac{\mathrm{w}}{\mathrm{M}}=\frac{10}{233}=0.0429$
$\therefore$ No. of moles of Na₂SO₄ needed $=\frac{\mathrm{M}\times \mathrm{V}}{1000}$
or $0.0429=\frac{5\times \mathrm{V}}{1000}\Rightarrow \mathrm{V}=8.58 \mathrm{mL}$