# An aqueous stock solution is #36.0%# HCl by mass and its density is #"1.18 g/mL"#. What volume of this solution is required to make #"1.00 L"# of #"1.75 mol/L"# #"HCl"_((aq))#? Give your answer in millilitres, accurate to three significant figures.

##### 1 Answer

#### Answer:

#### Explanation:

As you know, a **moles** of hydrochloric acid, the solute, **for every**

So right from the start, you know that your solution must contain **moles** of hydrochloric acid.

Use the compound's **molar mass** to convert this to *grams*

#1.75 color(red)(cancel(color(black)("moles HCl"))) * "36.46 g"/(1color(red)(cancel(color(black)("mole HCl")))) = "63.805 g"#

Now, you know that your stock solution is **by mass**, which implies that in order to have

You can thus say that in order for your target solution to contain *mass* of

#63.805 color(red)(cancel(color(black)("g HCl"))) * "100 g solution"/(36.0color(red)(cancel(color(black)("g HCl")))) = "177.24 g"#

Use the **density** of the stock solution to figure out the *volume* of the sample

#177.24 color(red)(cancel(color(black)("g solution"))) * "1 mL"/(1.18color(red)(cancel(color(black)("g solution")))) = color(darkgreen)(ul(color(black)("209 mL")))#

The answer is rounded to three **sig figs**.

So what you would do is take **add it** to enough water to get the total volume of the resulting solution to

Since you're working with concentrated hydrochloric acid, you should use an ice bath to cool the water before adding the stock solution to it.

You should also add the stock solution in *multiple steps* and stir the solution properly between each step.