QUESTIONS: I. If the molar extinction coefficient (c) of a compound is 75,000L m
ID: 956169 • Letter: Q
Question
QUESTIONS: I. If the molar extinction coefficient (c) of a compound is 75,000L mollem at 443 nm and the Abs (at 433 nm) = 0.73 find the concentration of that solution. (assume a path length of 1 .0 cm) = 73 C:A 26 2. If a student measures a sample that gives an Abs 1.93 at the mas is this data reliable? If not, then what might they do to improve their data? 3. If your stock solution has a concentration of 0.25 M and you want to make 250 mL. of a solution that has a concentration of 1.5 x 102M, how much stock solution will you need?Explanation / Answer
1. Absorbance = molar extinction coefficient x path length x concentration
Feed the given values,
concentration of the solution = 0.73/75000 x 1.0 = 9.73 x 10^-6 M
2. An absorbance of 1.93 is at a higher end of concentration limit. At high concentrations Beer Lambert's law is not applicable and data is not reliable. In order to improve the data, the concentration of the sample in solution must be diluted.
3. Using, M1V1 = M2V2
M1 and M2 are molar concentration of stock and final solutions, respectively
V1 and V2 are volumes of stock solution and final solution to be prepared, respectively
We get,
Volume of stock solution needed V1 = 1.5 x 10^-2 M x 250 ml/0.25 M = 15 ml