calculating volume percentages

Anyone?

I'm not sure I follow, but you are mixing 53 vol% and 57 vol% methanol to create 30 liters of 86 vol% methanol? This is impossible IMO, you could only get a solution in the interval (53 vol%,57 vol%). Never less never more.

>>8456848
agreed

>>8456848
I realize that and I made the same statement to a person trying to figure it out. Still, she insists that it's possible somehow using formula of some kind.
I don't think it's possible at all in real life, but maybe it's possible somehow mathematically using formula included. I'm not very good at math though when it comes to such things, I'm more of a chem guy, but I'm still interested if it's possible.
It's for someone trying to get inside pseudo-university which we have here in Finland.
I hope that I can get some clarity of this mess.

>>8457114
vol_MeOH = (%_a * vol_a) + (%_b * vol_b)
(where %_i is given as a volume fraction, ie 53% would be %_a = .53 in the formula)
vol_total = vol_a + vol_b
the limiting cases are vol_a = 0 and vol_b = 0

>I hope my english is understandable enough

>>8457130
Well, I'm not a native speaker after all.
I'm merely trying to figure out how could that kind of equation be possible, since I really can't see any way to make the desired solution.

>>8457288
>Well, I'm not a native speaker after all.
just to be clear, I wasn't mocking you for your command of English, rather I was sincerely wondering if there is some misunderstanding

b/c at this point it's either a misunderstanding w/in the thread or someone else (TA, professor, textbook author, w/e) fucked up when they wrote the problem