An impure sample of CaCO ₃ weighing 2.5 g is treated with 25 mL of an aqueous HCl solution. The HCl solution has a density of 1.2 g mL ⁻¹ and contains 7.3 % HCl by weight. After the reaction is complete, the excess HCl requires 10 mL of 2.0 M NaOH solution for complete neutralization. The percentage purity of the CaCO ₃ sample is ________. (Given: Molar mass of CaCO ₃ = 100 g mol ⁻¹ , HCl = 36.5 g mol ⁻¹ )

Step-by-step solution

Molarity of the HCl solution is given by M = 10 %( w/w ) d Molar mass M = 10 7.3 1.2 36.5 = 2.4 M Total millimoles of HCl added = 25 2.4 = 60 mmol Millimoles of excess HCl neutralized by NaOH = 10 2.0 = 20 mmol Millimoles of HCl reacted with CaCO ₃ = 60 - 20 = 40 mmol The reaction is: CaCO ₃ + 2 HCl CaCl ₂ + H ₂ O + CO ₂ Millimoles of CaCO ₃ reacted = 40 2 = 20 mmol = 0.02 mol Mass of pure CaCO ₃ = 0.02 100 = 2.0 g Percentage purity = 2.0 2.5 100 = 80 % Answer: 80