When 500 mL of 0.6 M hydrochloric acid is heated, its volume reduces to 250 mL and some HCl gas escapes. To determine the remaining concentration, a 25 mL portion of the final solution is titrated against 0.5 M NaOH solution. If 20 mL of the NaOH solution is required for complete neutralization, the mass of HCl gas that escaped is x 10⁻² g . The value of x is. (Molar mass of HCl is 36.5 g mol ⁻¹ )

Step-by-step solution

Moles of NaOH used in titration = M V = 0.5 0.020 = 0.01 mol Reaction: HCl + NaOH NaCl + H ₂ O Moles of HCl in the 25 mL aliquot = 0.01 mol Total moles of HCl in the remaining 250 mL solution = 0.01 250 25 = 0.10 mol Initial moles of HCl = 0.6 0.500 = 0.30 mol Moles of HCl escaped = 0.30 - 0.10 = 0.20 mol Mass of HCl escaped = 0.20 36.5 = 7.30 g Given mass = x 10⁻² g Therefore, x = 730 . Answer: 730