1000 g of a mixture of Na₂CO₃, Na₂SO₄ and NaOH for complete neutralization requires 511 g of HCl. The same mixture when reacted with excess of BaCl₂ solution, produced 466g of white precipitate of BaSO₄. Calculate mass % of NaOH in mixture.

If no of moles of Na₂CO₃ = a mole
No of moles of Na₂SO₄ = b mole
No of moles of NaOH = c mole
$\mathrm{a}\times 106+\mathrm{b}\times 142+\mathrm{c}\times 40=1000 \mathrm{gram}$
No of moles of HCl required for neutralization of mixture $2\mathrm{a}+\mathrm{c}=\frac{511}{36.5}=14$
No of moles of BaCl₂ = No of moles of BaSO₄
= No of moles of Na₂SO₄ (b)
$=\frac{466}{137+96}=\frac{466}{233}=2 \mathrm{moles}$
106a + 40c = 1000 – 284 = 716
80a + 40c = 14 × 40 = 560
$\mathrm{a}=\frac{156}{26}=6;\mathrm{c}=14-12=2$
So, wt. of NaOH = 2 × 40 = 80 gm
$\mathrm{\%}\text{ of }\mathrm{NaOH}=\frac{80}{1000}\times 100=8\mathrm{\%}$