# Question 6a626

Sep 19, 2016

$\text{5.7 mL}$

#### Explanation:

The idea here is that you can use the stock solution's mass by mass percent concentration and its density to figure out how many milliliters of solution would contain the number of moles of sulfuric acid needed to make the target solution.

So, you know that the target solution must have a volume of $\text{1 L}$ and a concentration of $\text{0.100 M}$. Use this info to calculate how many moles of acid it must contain

1 color(red)(cancel(color(black)("L solution"))) * ("0.100 moles H"_2"SO"_4)/(1color(red)(cancel(color(black)("L solution")))) = "0.100 moles H"_2"SO"_4

Now, you need to convert this to mass of sulfuric acid. To do that, use the compound's molar mass

0.100 color(red)(cancel(color(black)("moles H"_2"SO"_4))) * "98.08 g"/(1color(red)(cancel(color(black)("mole H"_2"SO"_4)))) = "9.808 g"

Next, use the stock solution's mass by mass percent concentration to find the mass of solution that would contain $\text{9.808 g}$ of sulfuric acid

9.808 color(red)(cancel(color(black)("g H"_2"SO"_4))) * overbrace("100 g stock solution"/(94color(red)(cancel(color(black)("g H"_2"SO"_4)))))^(color(blue)("= 94% m/m")) = "10.434 g stock solution"

Finally, use the solution's density to convert this to cubic centimeters

10.434 color(red)(cancel(color(black)("g stock solution"))) * "1 cm"^3/(1.831color(red)(cancel(color(black)("g stock solution")))) = "5.7 cm"^3#

As you know, $\text{1 cm"^3 = "1 mL}$, and so the answer will be

$\textcolor{g r e e n}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{volume of stock solution " = " 5.7 mL}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

I'll leave the answer rounded to two sig figs.