10 g of NaOH is dissolved in 10 ml of 5M HCl solution, excess of HCl is neutralized by 5ml of  2M $\text{Ba}{\left(\text{OH}\right)}_2$ solution. Percentage purity of caustic soda is

Equivalent of HCl = 5 ×10 ml = 50 milliequivalents Equivalent of Ba(OH)2 = 4 ×5 ml = 20 milliequivalents Acid used = 50 −20 = 30 milliequivalents

NaOH w.t = 30 × 40 = 1200 mg = 1.2 g Percentage = (1.2/10) × 100 = 12%