1.190x10^-1 m/l
First, calculate how many moles of NaOH was used.
0.03430 l * 8.670x10^-2 mol/l = 2.97381x10^-3 mol
So we used 2.97381x10^-3 moles of NaOH, which means we had 2.97381x10^-3 moles of HCl in the original sample. So the molarity of the original sample is the number of moles divided by the volume.
2.97381x10^-3 mol / 0.02500 = 0.1189524 m
Rounding to 4 significant figures gives 0.1190 or 1.190x10^-1 m
