The balanced chemical equation for the reaction between CaCO3 and HCl is:

CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)

From the equation, we can see that 1 mole of CaCO3 reacts with 2 moles of HCl. The molar mass of CaCO3 is 100.09 g/mol, so 1.25 grams of CaCO3 is equal to 1.25 / 100.09 = 0.0125 moles of CaCO3.

Since 0.0125 moles of CaCO3 reacts with 25.5 mL of HCl solution, the molarity of the HCl solution can be calculated as follows:

Molarity = moles of solute / volume of solution in liters

Molarity = 0.0125 moles / (25.5 mL / 1000) L

Molarity = 0.0125 moles / 0.0255 L

Molarity = 0.489 M

Therefore, the molarity of the HCl solution is 0.489 M.