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
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.
To find the concentration of the diluted solution, we need to calculate the amount of ASA (acetylsalicylic acid) present in the 3.00 mL solution and then dilute it to a total volume of 100 mL.
Here's how you can calculate it step by step:
Step 1: Calculate the mass of ASA in the 0.475 g aspirin tablet.
Mass of ASA = 0.475 g x 0.682 (mass fraction of ASA)
Mass of ASA = 0.32435 g
Step 2: Calculate the moles of ASA.
Moles of ASA = Mass of ASA / Molar mass of ASA
Molar mass of ASA = 180.16 g/mol
Moles of ASA = 0.32435 g / 180.16 g/mol
Moles of ASA = 0.0018002 mol
Step 3: Calculate the concentration of the diluted solution.
Moles of ASA in the 3.00 mL solution = Moles of ASA x (3.00 mL / 250 mL)
Moles of ASA in the 3.00 mL solution = 0.0018002 mol x (3.00 mL / 250 mL)
Moles of ASA in the 3.00 mL solution = 0.00002161 mol
Concentration of the diluted solution = Moles of ASA in the 3.00 mL solution / 0.100 L
Concentration of the diluted solution = 0.00002161 mol / 0.100 L
Concentration of the diluted solution = 0.2161 mol/L
So, the concentration of the diluted solution is 0.2161 mol/L.