An aspirin tablet weighing 0.400 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 concn of what in the diluted solution? the original sample or the ASA in the original sample? And what kind of units for the concn?

The original sample is
(0.400 g/250 mL) x (3 mL/100 mL) = ?? g/mL

If you want the concn ASA, you start with 0.4 x 0.682 = ?? and use that number in place of the 0.400 g.

If you want molarity, change the first solution to molarity.
0.400 x 0.682 x (1 mol/180.16 g) x 1000 mL/250 mL then the dilution is x (3 mL/100 mL) = ?? M.
Check my work.

Your work is correct, DrBob222. Thanks for the help

so whats the actual answer for the concentration? i keep trying and im getting the wrong answer.

To determine the concentration of the diluted solution, we need to calculate the amount of ASA present in the 3.00 mL solution pipetted from the tablet solution.

1. Start by calculating the mass of ASA in the original tablet:

Mass of ASA = Mass of tablet * %ASA by mass
= 0.400 g * 0.682
= 0.2728 g

2. Calculate the number of moles of ASA:

Number of moles = Mass / Molar mass
= 0.2728 g / 180.16 g/mol
= 0.00151 mol

3. Using the dilution formula C1V1 = C2V2, where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume, we can calculate the concentration of the diluted solution.

C1 = Concentration of the original solution = Number of moles / Volume of the original solution
= 0.00151 mol / 0.250 L
= 0.00604 mol/L

V1 = Volume of solution pipetted = 3.00 mL = 0.003 L

C2 = Concentration of the diluted solution
V2 = Volume of the diluted solution = 100 mL = 0.100 L

Substituting the values into the dilution formula, we have:

0.00604 mol/L * 0.003 L = C2 * 0.100 L

C2 = (0.00604 mol/L * 0.003 L) / 0.100 L
= 0.00001812 mol/L

Therefore, the concentration of the diluted solution is 0.00001812 mol/L or 1.812 * 10^-5 mol/L.