Question

Question #03d29

Answer 1

Here's what I got.

Explanation:

The first thing to do here is use the volume by volume percent concentration, `"%v/v"`, of the target solution to determine how many liters of solute, which in your case is sulfuric acid, `"H"_2"SO"_4`, it must contain.

A `"29% v/v"` sulfuric acid solution will contain `"29 L"` of sulfuric acid for every `"100 L"` of solution, which means that your solution must contain

`600 color(red)(cancel(color(black)("L solution"))) * ("29 L H"_2"SO"_4)/(100color(red)(cancel(color(black)("L solution")))) = "174 L H"_2"SO"_4`

Now, let's assume the `x` represents the volume of the `"70% v/v"` sulfuric acid solution and `y` represents the volume of the `"25% v/v"` sulfuric acid solution.

The first equation that you can write here will be

`x + y = "600 L"" " " "color(orange)((1))`

This simply uses the fact that the two solutions must be mixed together to form a total volume of `"600 L"`.

Now, use the given percent concentrations to figure out how many liters of sulfuric acid you'd get in those two solutions

`xcolor(white)(a)color(red)(cancel(color(black)("L solution"))) * ("70 L H"_2"SO"_4)/(100color(red)(cancel(color(black)("L solution")))) = 7/10xcolor(white)(a)"L H"_2"SO"_4`

`ycolor(white)(a)color(red)(cancel(color(black)("L solution"))) * ("25 L H"_2"SO"_4)/(100color(red)(cancel(color(black)("L solution")))) = 1/4ycolor(white)(a)"L H"_2"SO"_4`

This means that you can write

`7/10x + 1/4y = "174 L"" " " "color(orange)((2))`

This equation describes the fact that the amount of sulfuric acid you get from the two solutions you're mixing must add up to give `"174 L"`.

Use equation `color(orange)((1))` to get

`x = 600 - y`

Plug this into the second equation to get

`7/10(600 -y) + y/4 = 174`

`420 - 7/10y + y/4 = 174`

`-9/20y = -246 implies y = 547`

This means that you have

`x = 600 - 547 = 53`

I'll leave the answers as

`"volume of 70% solution" = color(green)(|bar(ul(color(white)(a/a)"50 L"color(white)(a/a)|)))`

`"volume of 25% solution" = color(green)(|bar(ul(color(white)(a/a)"550 L"color(white)(a/a)|)))`

Finally, the result makes sense because the concentration of the target solution is much closer to the concentration of the `25%` solution, which implies that you'd need a lot more of this solution than of the `70%` one.