$$HCl + NaOH \rightarrow NaCl + H_2O$$
From the equation, we can see that 1 mole of HCl reacts with 1 mole of NaOH. Therefore, the number of moles of HCl in the 10.0 mL sample can be calculated as follows:
$$moles of HCl = (26.13 mL)(0.108 M) = 2.812 \times 10^{-3} moles$$
The molar concentration of HCl in the original sample can be calculated by dividing the number of moles of HCl by the volume of the sample in liters:
$$molarity of HCl = \frac{2.812 \times 10^{-3} moles}{10.0 \times 10^{-3} L} = 0.2812 M$$
Therefore, the molar concentration of HCl in the original sample is 0.2812 M.
