The organic chemical aniline, C6H5NH2, was to be determined instrumentally. 5.0
ID: 978939 • Letter: T
Question
The organic chemical aniline, C6H5NH2, was to be determined instrumentally. 5.0 mg of pure aniline was dissolved in 1.00 L of water. The instrument analyzed this solution, and gave an output reading of 0.655. Assume the instrument responds linearly. An impure unknown sample of aniline weighing 0.200 g was then dissolved in 500.0 ml of water. A 25.00 ml aliquot of this was then diluted to 250.0 ml and mixed well. A 10.0 ml aliquot of this was then diluted to 100.0 ml and measured on the same instrument under the same conditions, giving a reading of 0.425 units. What was the % purity of the unknown alinine sample?
Explanation / Answer
5 mg aniline in 1 L gave reading of 0.655
in 100 ml amount of aniline present would be 0.5 mg with reading 0.0655
0.425 will correspond to an aniline concentration in 100 ml = 3.24 mg
aniline present in 10 ml = 3.24 mg
aniline present in 250 ml = 25 x 3.24 = 81.0 mg
aniline present in 25 ml = 81.0 mg
aniline present in 500 ml = 81 x 2 = 162 mg
% purity of sample = (0.162/0.200) x 100 = 81.0%