chemistry
posted by jared on .
An aspirin tablet weighing 0.548 g has been analyzed and contains 68.2 % ASA (180.16 g/mol) by mass. A student dissolved the tablet in hot NaOH and the cooled solution was diluted with DI water to the mark in a 250 mL volumetric flask. Exactly 3.00 mL
of the solution was pipetted into a 100 mL volumetric flask and diluted to the mark with FeCl3 solution.
The concentration of the diluted solution is
________M.
mass aspirin = 0.548*0.628 = ??
??/180.16 = xx mols aspirin = 0.00191 mols. Check my arithmetic.
diluted to 250 so 0.00191 mols/250 = xx mols/mL.
Take 3 mL of that solution and make to mark of 100, so we now have xx mols/mL x 3 mL/100 mL = yy mols/mL.
We want mols /L so multiply yy mols/mL *1000 to obtain mols/L = molarity.

take the amount of grams (.548) and multiply it by .628
You then get .37373 (this is the concentration of ASA that the problem is focusing on)
So you divide .37373 by 180.16 to figure out the amount of moles= .002075, divide this number by .250L to get your Molarity, your concentration= .008299mols/L
then you plug this in the formula
M1 x V1 = M2 x V2
M1= .008299
V1= .003 L (the amount of the solution you use)
M2= ? (this is our variable that we are trying to find)
V2= .100 L
your M2 comes out to be .000249mols/L
ANd that's the answer!!! (: