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.