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???
ive tried it a couple times but i cant figure it out....
chem - DrBob222, Monday, April 18, 2011 at 4:42pm
You mustn't let these problems confuse you. Stick to what is going on and don't get distracted.
0.475 g of 68.2% ASA==> 250 mL.
Then 3.00 ==> 100 mL.
0.475 x 0.682 x (1 mol/180.18) = ?? moles ASA placed in the 250 mL volumetric flask; therefore, the concn of ASA in that 250 mL flask is just ?? moles/0.250 mL = xxM ASA. Right?
Then we take 3.00 mL and dilute it to 100; therefore, the concn in the 100 mL flask is
xxM x (3.00/100) = zzM.
chem - CHEMCHEM, Monday, April 18, 2011 at 8:34pm
OMMMGGGGG i freaking love you bob thank youuu <333333