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A 11 mL sample of a solution of AlBr3 was diluted with water to 23 mL. A 15 mL sample of the dilute solution was found to contain 0.045 moles of Br-. What was the concentration of AlBr3 in the original undiluted solution?

answer is 2.09 but i cant get it.

0.045 mols = 45 millimoles which I like better; it keeps all those zeros out of the way.
45 mmoles Br^- x 1/3 = mmoles AlBr3 since there are 3 Br^-/molecule AlBr3. That's 15 mmoles in the 15 mL.

That was the aliquot from 23 mL; therefore 15 mmoles x 23/15 was mmoles in the 23 mL sample. 15 x 23/15 = 23 mmoles (you don't even need a calculator so far).

So we had 23 mmoles in the original 11 mL. M = mmoles/mL = 23/11 = 2.09M

Kate, I think I worked a problem for you earlier that looked a lot like this. If so I must have not read it carefully, for I don't think I approached it this way. If so be sure and check that answer versus this problem.

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