# Question #6a626

##### 1 Answer

#### 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 **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

#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,

#color(green)(bar(ul(|color(white)(a/a)color(black)("volume of stock solution " = " 5.7 mL")color(white)(a/a)|)))#

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