The reaction is
[tex]\text{ KOH + HCl }\rightarrow\text{ KCl + H}_2\text{O}[/tex]moles of acid
[tex]\text{ 19.5 mL KOH }\times\frac{1\text{ L}}{\text{ 1000 mL}}\times\frac{\text{ 0.7 mol KOH}}{1\text{ L KOH}}\times\frac{1\text{ mol HCl}}{1\text{ mol KOH}}=\text{ 0.01365 mol HCl}[/tex]Molarity of acid
[tex]\text{ M = }\frac{\text{ moles of acid}}{\text{ Volume of acid}}=\text{ }\frac{\text{ 0.01365 mol}}{\text{ 0.0268 L}}=\text{ 0.51 M}[/tex]The answer is: the molarity of the acid solution is 0.51 M