# Question aadce

Jul 30, 2017

$\text{800 mL}$

#### Explanation:

For starters, you should know that an $\text{80% v/v}$ methanol solution will contain $\text{80 mL}$ of methanol, the solute, for every $\text{100 mL}$ of solution.

If you take $x$ $\text{mL}$ to be the volume of pure methanol that must be added to the $\text{200 mL}$ sample of water, you can say that the total volume of the resulting solution will be

$x \textcolor{w h i t e}{.} \text{mL" + "200 mL" = (200 + x)color(white)(.)"mL}$

Since you know that this solution must contain $\text{80 mL}$ of methanol for every $\text{100 mL}$ 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")))#

$100 \cdot \frac{x}{200 + x} = 80$

Solve for $x$ to get

$100 \cdot x = 80 \cdot \left(200 + x\right)$

$100 \cdot x = 16000 + 80 x$

$20 x = 16000 \implies x = \frac{16000}{20} = 800$

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

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