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 ______$$\times 10^{-1}$$. (nearest integer)
Given : Molar mass of HCl and $$Ba(OH)_{2}$$ are 36.5 and 171.0 g $$mol^{-1}$$ respectively.

We need to find the mass $$x$$ (in mg) of pure HCl that reacts completely with 25.0 mL of 0.1 M $$Ba(OH)_2$$.

The balanced equation is $$2HCl + Ba(OH)_2 \rightarrow BaCl_2 + 2H_2O$$ and the mole ratio is 2 mol HCl per 1 mol $$Ba(OH)_2$$.

The moles of $$Ba(OH)_2$$ are calculated as Molarity $$\times$$ Volume = $$0.1 \times 0.025 = 0.0025$$ mol, which equals 2.5 mmol.

The moles of HCl required are $$2 \times 2.5 = 5$$ mmol, or 0.005 mol.

The mass of HCl is then moles $$\times$$ molar mass = $$0.005 \times 36.5 = 0.1825$$ g, which equals 182.5 mg.

Expressing in the required form gives $$x = 182.5$$, so $$x \times 10^{-1} = 1825$$.

The answer is 1825.