An aspirin tablet weighing 0.400 g has been analyzed and contains 68.2% acetylsalicylic acid (asa) (180.16 g/mol) by mass. a student dissolved the tablet in hot naoh and the cooled solution was diluted with deionized 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. calculate the concentration of asa in the diluted solution.

0.00018M

Explanation:

Mass of the aspirin tablet =0.4g.

% by mass of ASA=68.2

mass of in the tablet=68.2/100×0.4g

=0.2748g

mass conc of ASA=mass of ASA/Volume of diluent(250ml)

=0.2748/0.25

=1.0992g/dm3

molar conc=mass conc/molar mass

=1.0992/180.16

=0.006M

Conc of the stock solution C1=0.006M

Initial volume for making the diluted solution(V1)=3ml

final volume of the diluted solution=100ml

from dilution principle

no of mole=conc × volume

initial conc(C1)×initial volume(V1)=final conc(C2)×final volume(V2)

C2=0.006M×3ml/100ml

=0.00018M

the correct answer would be d. hope this !