Question #ba500

The problem provides you with the volume of solute, which in your case is ethanol, `"CH"_3"CH"_2"OH"`, and the volume of solvent, which in your case is water, so right from the start you know that you're looking for the solution's volume by volume percent concentration, `"% v/v"`.

The solution's volume by volume percent concentration tells you how many cubic centimeters of solute you get for every `"100 cm"^3` of solution. The first thing to do here will be to figure out the total volume of the solution.

`V_"total" = V_"solute" + V_"solvent"`

In your case, you have

`V_"total" = "40 cm"^3 + "160 cm"^3 = "200 cm"^3`

Now, you know that you have `"40 cm"^3` of ethanol in `"200 cm"^3` of solution, which means that `"100 cm"^3` of solution will contain

`100 color(red)(cancel(color(black)("cm"^3color(white)(a)"solution"))) * ("40 cm"^3color(white)(a)"CH"_3"CH"_2"OH")/(200color(red)(cancel(color(black)("cm"^3color(white)(a)"solution")))) = "20 cm"^3color(white)(a)"CH"_3"CH"_2"OH"`

So, if `"100 cm"^3` of solution contain `"20 cm"^3` of ethanol, it follows that the solution's volume by volume percent concentration is

`"% v/v" = color(green)(|bar(ul(color(white)(a/a)color(black)("20% CH"_3"CH"_2"OH")color(white)(a/a)|)))`