What is the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl (molar mass = 36.5 g) into enough water to make 26.8 mL of solution?

The first thing to check here is how much hydrogen chloride, `"HCl"`, can you dissolve in water at room temperature. This will help you make sure that all the mass of hydrogen chloride will actually dissolve.

At room temperature, hydrogen chloride has a solubility of about `"720 g/L"`, which means that your `"26.8-mL"` sample will hold as much as

`26.8color(red)(cancel(color(black)("mL"))) * "720 g HCl"/(1000color(red)(cancel(color(black)("mL")))) = "19.3 g HCl"`

Now that you know that all the hydrogen achloride will actually dissolve, you can use its molar mass to determine how many moles you have in the `"1.56-g"` sample

`1.56color(red)(cancel(color(black)("g"))) * "1 mole HCl"/(36.5color(red)(cancel(color(black)("g")))) = "0.04274 moles HCl"`

Now, molarity is defined as moles of solution, which in your case is hydrogen chloride, divided by liters of solution.

`color(blue)("molarity" = "moles of solute"/"liters of solution")`

In your case, the molarity of the resulting solution will be

`c = "0.04274 moles"/(26.8 * 10^(-3)"L") = "1.5948 M"`

Rounded to three sig figs, the answer will be

`c = color(green)("1.59 M")`