An aqueous solution is 40.0% by mass hydrochloric acid, HCl, and has a density of 1.20 g/mL. What is the molarity of hydrochloric acid in the solution?

The first thing to do here is pick a sample of this solution.

Since you're going to find its molarity, you can make the calculations easier by picking a #"1.00-L"# sample. Keep in mind ,the answer will be the same regardless of the volume you pick.

So, use the density of the solution to determine how many grams you get in that respective volume

#1.00color(red)(cancel(color(black)("L"))) * (1000color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1.20 g"/(1color(red)(cancel(color(black)("mL")))) = "1200 g"#

Now, you that the percent concentration by mass for this solution is #"40.0% HCl"#. This means that you get #"40.0 g"# of hydrochloric acid for every #"100.0 g"# of solution.

The mass of hydrochloric acid you get in that sample will thus be

#1200color(red)(cancel(color(black)("g solution"))) * "40.0 g HCl"/(100color(red)(cancel(color(black)("g solution")))) = "480 g HCl"#

Use hydrochloric acid's molar mass to determine how many moles you have in that many grams

#480color(red)(cancel(color(black)("g"))) * "1 mole HCl"/(36.46color(red)(cancel(color(black)("g")))) = "13.17 moles HCl"#

The molarity of the solution will thus be - remember that we started with a #"1.00-L"# sample

#c = n/V#

#c = "13.17 moles"/"1.00 L" = color(green)("13.2 M") -># rounded to three sig figs

SIDE NOTE This is very close to the maximum possible concentration for hydrochloric acid in aqueous solution. At this concentration, the solution is actually fuming at a significant evaporation rate.