Changing the Concentration of a SolutionDate: 05/30/2002 at 20:31:37 From: Makeia Gay Subject: I need help I don't understand how to do mixture problems. For example, a chemist has 6 liters of a 25% alcohol solution. How much alcohol must he add so that the resulting solution contains 50% alcohol? Date: 05/31/2002 at 14:25:22 From: Doctor Ian Subject: Re: I need help Hi Makeia, The way I normally do these is with pictures. Suppose I have 6 liters of 25% alcohol. That would look like +---+ +---+ +---+ +---+ +---+ +---+ | | | | | | | | | | | | +---+ +---+ +---+ +---+ +---+ +---+ | | | | | | | | | | | | +---+ +---+ +---+ +---+ +---+ +---+ | | | | | | | | | | | | +---+ +---+ +---+ +---+ +---+ +---+ | A | | A | | A | | A | | A | | A | +---+ +---+ +---+ +---+ +---+ +---+ Each 'tube' represents a liter, and each liter is 1/4 alcohol. Does that make sense? Now, suppose he's going to add pure alcohol until he ends up with a 50% solution. How much does he have to add? The picture tells us that if we add two more 'rows' of A's at the bottom, we'd have a 50% solution: +---+ +---+ +---+ +---+ +---+ +---+ | | | | | | | | | | | | +---+ +---+ +---+ +---+ +---+ +---+ | | | | | | | | | | | | +---+ +---+ +---+ +---+ +---+ +---+ | | | | | | | | | | | | +---+ +---+ +---+ +---+ +---+ +---+ | A | | A | | A | | A | | A | | A | +---+ +---+ +---+ +---+ +---+ +---+ +---+ +---+ +---+ +---+ +---+ +---+ | A | | A | | A | | A | | A | | A | +---+ +---+ +---+ +---+ +---+ +---+ | A | | A | | A | | A | | A | | A | +---+ +---+ +---+ +---+ +---+ +---+ So how much alcohol would that be? Each 'A' is 1/4 of a liter, and there are 12 of them; so he has to add 12/4 liters, or 3 liters. I hope this helps. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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