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.
Answer
1825
Solution
2HCl + Ba(OH)2 BaCl2 + 2H2O
Moles of Ba(OH)2 = 0.1 25.01000 = 2.5 10-3 mol
moles of HCl = 2 2.5 10-3 = 5 10-3 mol
mass of HCl = 5 10-3 36.5 = 0.1825 g = 182.5 mg
x = 182.5 = 1825 10-1
Answer: 1825