Question #aadce

For starters, you should know that an `"80% v/v"` methanol solution will contain `"80 mL"` of methanol, the solute, for every `"100 mL"` of solution.

If you take `x` `"mL"` to be the volume of pure methanol that must be added to the `"200 mL"` sample of water, you can say that the total volume of the resulting solution will be

`x color(white)(.)"mL" + "200 mL" = (200 + x)color(white)(.)"mL"`

Since you know that this solution must contain `"80 mL"` of methanol for every `"100 mL"` of solution, you can say that you will have

`100 color(red)(cancel(color(black)("mL solution"))) * (x color(white)(.)"mL methanol")/((200 + x) color(red)(cancel(color(black)("mL solution")))) = "80 mL methanol"`

This will get you

`100 * (xcolor(red)(cancel(color(black)("mL methanol"))))/(200 + x) = 80 color(red)(cancel(color(black)("mL methanol")))`

`100 * x/(200 + x) = 80`

Solve for `x` to get

`100 * x = 80 * (200 + x)`

`100 * x = 16000 + 80x`

`20x = 16000 implies x= 16000/20 = 800`

You can thus say that if you add `"800 mL"` of methanol to `"200 mL"` of water, you will end up with a solution that is `"80% v/v"` methanol.

The answer must be rounded to one significant figure, the number of sig figs you have for the volume of water.