Changing the Concentration of a Solution

Date: 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/