# 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?

Nov 11, 2015

$\text{1.59 M}$

#### Explanation:

The first thing to check here is how much hydrogen chloride, $\text{HCl}$, 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 $\text{720 g/L}$, which means that your $\text{26.8-mL}$ 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 $\text{1.56-g}$ 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.

$\textcolor{b l u e}{\text{molarity" = "moles of solute"/"liters of solution}}$

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

$c = \text{0.04274 moles"/(26.8 * 10^(-3)"L") = "1.5948 M}$

Rounded to three sig figs, the answer will be

$c = \textcolor{g r e e n}{\text{1.59 M}}$