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?
YES
To determine the concentration of the diluted solution, we need to gather all the necessary information and perform a series of calculations.
Given:
- Mass of aspirin tablet = 0.502 g
- ASA content by mass = 68.2%
- ASA molar mass = 180.16 g/mol
- Volume of the final solution = 100 mL
First, we need to calculate the mass of ASA (acetylsalicylic acid) present in the tablet:
Mass of ASA = Mass of tablet * ASA content by mass
Mass of ASA = 0.502 g * 0.682
Mass of ASA = 0.342 g
Next, we need to calculate the number of moles of ASA:
Number of moles of ASA = Mass of ASA / Molar mass of ASA
Number of moles of ASA = 0.342 g / 180.16 g/mol
Number of moles of ASA = 0.0019 mol
Now, we need to calculate the concentration of the solution after dilution:
Concentration = Number of moles / Volume (in liters)
Volume (in liters) = Volume (in mL) / 1000
Concentration = 0.0019 mol / (3.00 mL / 1000)
Concentration = 0.0019 mol / 0.003 L
Concentration = 0.633 M
Therefore, the concentration of the diluted solution is 0.633 M.