# 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.

Sep 27, 2017

$\text{209 mL}$

#### Explanation:

As you know, a ${\text{1.75-mol L}}^{- 1}$ hydrochloric acid solution will contain $1.75$ moles of hydrochloric acid, the solute, for every $\text{1 L}$ of solution.

So right from the start, you know that your solution must contain $1.75$ 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 36.0% hydrochloric acid by mass, which implies that in order to have $\text{36.0 g}$ of hydrochloric acid, you need $\text{100 g}$ of this solution.

You can thus say that in order for your target solution to contain $\text{63.805 g}$ of hydrochloric acid, the sample you take from the stock solution must have a 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

$177.24 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g solution"))) * "1 mL"/(1.18color(red)(cancel(color(black)("g solution")))) = color(darkgreen)(ul(color(black)("209 mL}}}}$

The answer is rounded to three sig figs.

So what you would do is take $\text{209 mL}$ of the stock solution and add it to enough water to get the total volume of the resulting solution to $\text{1.00 L}$.

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.