The balanced chemical reaction of the HCl reacting with `BaOH_2` is:

`2HCl + BaOH_2 -> BaCl_2 + 2H_2O`

In reactions involving titration first determine the number of H+ and OH- ions liberated by a molecule of the acid and base respectively. Here, one molecule of HCl ionizes to give one ion of H+ and one molecule of BaOH2 forms 2 ions of OH-

142 mL of 0.25 HCl releases 0.142*0.25 = 0.0355 moles of H+. These react with the OH- ions released by the `BaOH_2` to produce water. 0.0355 moles of OH- ions are released by 0.01775 moles of `BaOH_2` . If the molarity of `BaOH_2` is M, M*0.240 = 0.01775

=> M = 0.0739

The molarity of `BaOH_2` is approximately 0.0739