x mg of pure HCl was used to make an aqueous solution. 25.0 mL of 0.1 M Ba ( OH )_ 2 solution is used when the HCl solution was titrated against it. The numerical value of x is \_\_\_\_ 10^ -1 . (Nearest integer) Given : Molar mass of HCl and Ba ( OH )_ 2 are 36.5 and 171.0 ~g ~mol ^ -1 respectively.
A. A
The reaction between HCl and Ba(OH)₂ is: 2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O Moles of Ba(OH)₂ used: n = 0.1 M × 0.025 L = 0.0025 mol From stoichiometry: moles of HCl = 2 × 0.0025 = 0.005 mol Mass of HCl = 0.005 mol × 36.5 g/mol = 0.1825 g = 182.5 mg Expressing as x = ___ × 10⁻¹: 182.5 = 1825 × 10⁻¹ The numerical value is 1825 (or 1.825 × 10² mg when expressed with standard scientific notation)
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