Question
An aspirin tablet weighing 0.475 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
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
Answers
What kind of concn? M, m, %, g/mL, what? If M, then
g ASA in the tablet = 0.475 g x 0.682 = ??g ASA
??g/180.16 = xx moles ASA
M = xx moles/0.250L = yy M
Then yyM x (3/100) = zz M as the final answer.
g ASA in the tablet = 0.475 g x 0.682 = ??g ASA
??g/180.16 = xx moles ASA
M = xx moles/0.250L = yy M
Then yyM x (3/100) = zz M as the final answer.