Question

A solution contains 0.00750M of a first compound with an extinction coefficient of 22.4cm−1M−1. A second compound is added to the solution at a concentration of 0.0260M. The absorbance of the resulting solution is 0.324 with a path length of 1.00cm. What is the extinction coefficient for the second compound in the solution? Round the answer to three significant figures. Give the answer to three significant figures.

A solution contains 0.00750 M of a first compound with an extinction coefficient of 22.4 cmM.Asecond compound is added to the solution at a concentration of 0.0260 M. The absorbance of the resulting solution is 0.324 with a path length of 1.00 cm. What is the extinction coefficient for the second compound in the solution? Round the answer to three significant figures. Give the answer to three significant figures. Provide your answer below cm 'M-1

Answer 1

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According to Beer's - lambert's law A = epsilon C l A = Absorbance epsilon = molar extinction coefficient C = concentration l = path length For compound (1) epsilon = 22.4 cm^-1 m^-1 l = 1.0cm c = 0.0075 M then A = (22.4 cm^-1 m^-1) (0.0075 m) (1.0 cm) A = 0.168 For mixture of compound Absorbance is 0.324 so for second compound = 0.324 - 0.168 = 0.156 and concentration = 0.026 M therefore 0.156 = epsilon (0.026 M) (1.00 cm) epsilon 0.156/(0.026 M) (1.0 cm) epsilon 6 m^-1 cm^-1