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Posted by on Monday, January 28, 2008 at 4:37am.

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

  • chemistry - , Monday, January 28, 2008 at 9:27am

    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.

  • chemistry - , Monday, January 28, 2008 at 1:23pm

    thank you i got it

  • chemistry - , Sunday, January 25, 2009 at 6:38pm

    if I'm following this correctly, the answer works out to 2.28E-4 correct?

  • chemistry - , Saturday, April 16, 2011 at 3:31pm


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