# Concentrated hydrochloric acid is usually available at a concentration of 37.7 percent by mass. The density of the solution is 1.19 g/mL. What is its molar concentration?

Nov 30, 2015

$\text{12.3 M}$

#### Explanation:

As you know, molarity is defined as moles of solute, which in your case will be hydrochloric acid, $\text{HCl}$, divided by liters of solution.

This means that you can make the calculations easier by picking a $\text{1.00-L}$ sample of this concentrated hydrochloric acid solution. In this particular case, the molarity of the solution will be equal to the number of moles of solute.

So, use the solution's density to determine what the mass of this $\text{1.00-L}$ sample would be

1.00color(red)(cancel(color(black)("L"))) * (1000color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1.19 g"/(1color(red)(cancel(color(black)("mL")))) = "1190 g"

Now, you know that the percent concentration by mass of this solution is equal to 37.7%. This means that you get $\text{37.7 g}$ of hydrochloric acid for every $\text{100.0 g}$ of solution.

In this case, your sample would contain

1190 color(red)(cancel(color(black)("g solution"))) * "37.7 g HCl"/(100color(red)(cancel(color(black)("g solution")))) = "448.63 g HCl"

All you have to do now is use hydrochloric acid's molar mass to find how many moles would be present in $\text{448.63 g}$.

448.63 color(red)(cancel(color(black)("g"))) * "1 mole HCL"/(36.46color(red)(cancel(color(black)("g")))) = "12.304 moles HCl"

Therefore, the solution's molarity will be

$\textcolor{b l u e}{c = \frac{n}{V}}$

c = "12.304 moles"/"1.00 L" = color(green)("12.3 M")

The answer is rounded to three sig figs.