We're asked to find the molar concentration (molarity) of the #"HCl"# solution, given some titration measurements.

Let's first write the chemical equation for this neutralization reaction:

#"NaOH"(aq) + "HCl" (aq) rarr "NaCl" (aq) + "H"_2"O" (l)#

Since we're given the molarity and volume of the #"NaOH"# solution used, we can calculate the moles be using the molarity equation:

#"mol solute" = ("molarity")("L soln")#

#= (0.1"mol"/(cancel("L")))(0.0332cancel("L")) = color(red)(0.00332# #color(red)("mol NaOH"#

Since all the coefficients in the chemical equation are #1#, the relative number of moles of #"HCl"# used is also #color(red)(0.00332# #color(red)("mol"#.

Finally, let's use the molarity equation again to find the molarity of the #"HCl"# solution (given the volume of #"NaOH soln"# is #40# #"mL"#, which must be in **liters** when using molarity equations):

#"molarity" = "mol solute"/"L soln"#

#= (color(red)(0.00332)color(white)(l)color(red)("mol HCl"))/(0.040color(white)(l)"L") = color(blue)(0.083M#