posted by Anonymous .
An aspirin tablet weighing 0.502 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
DrWLS worked this problem yesterday.
0.502 g x 0.682 = g ASA in tablet.
g ASA in tablet/molar mass = mols ASA in tablet.
mols ASA in tablet/0.250 L = xx molarity in the 250 mL flask.
Dilution was 3:100; therefore, the new solution is xx M x 3/100 = yy M in the 100 mL flask.
Post your work if you get stuck.
thank you i got it
if I'm following this correctly, the answer works out to 2.28E-4 correct?