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July 26, 2014

Homework Help: chemistry

Posted by jared on Thursday, May 10, 2007 at 10:08pm.

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.

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