[schmitgreg]

Hey, I have been practicing molarity, and I ran across an example problem on about.com and I tried it on paper, but I got stuck. there are two parts, and here is the link:
http://chemistry.about.com/library/weekly/bl062503b.htm

Part a was easy, all I had to do was setup the problem to find how many grams of BaCl₂ I needed to prepare a .10 M solution of BaCl₂. My answer was 520.581 g BaCl₂ and sure enough, I was right. So, 2.5 mol BaCl₂ / 25 liters gave .1 M BaCl₂.

Part b seemed easy, but now I am not so sure. The problem was to find out how much solvent (water) you need to dilute the prepared solution in part 1 to .02 M. This is what I did:

2.5 mol / X liters = 2 mol / 100 liters
I did this because 2 mol / 100 liters is a .02 M solution and I needed to find out how much solvent I need.
By solving for X I got X = 125
I then saw that I had 25 liters of solution already from part a so I subtracted that from my 125 liters and got 100 liters of solvent were needed to dilute the solution.

I was wrong. The answer listed was .2 liters or 200 cm³

I'm not completely sure what I did wrong, but can someone shed some light? Thanks!

[Padfoot]

The problem was to find out how much solvent (water) you need to dilute the prepared solution in part 1 to .02 M.

No it wasn't.
It was to determine what volume of your solution contains 0.02mol.
They use v=n/c. 

If it were a dilution problem, your eqn would still be wrong - it implies concentration is constant.  If it were a dilution problem you would use c1v1=c2v2 - this implies n is constant