Erioglucine is a blue-colored dye that absorbs its complementary color, red, in aqueous solution around 645 nm. Unfortunately, the local distilled water supply used to make solutions is consistently contaminated with a trace amount of a metal cation that also absorbs 645 nm light. Suppose then, a control sample of 0.0552 M erioglucine (aq) has an absorbance of 0.331 and that a distilled water sample in a similar cuvette has an absorbance of 0.019. Determine the concentration of an erioglucine (aq) sample that has an absorbance of 0.217. 10 pts. The answer is .0350 but I wanted to see how to get that answer. You have to subtract .019 from the absorbance of the control and the sample

For the control experiment: of aqueous solution of erioglaucine has absorbance of

From Lambert-Beer's law we know:

Here; e is the molar absorptivity coefficient of erioglaucine

l = length of cuvette in which the solution is taken =

A sorbance by the erioglaucine = total absorbance - absorbance by distilled

So; by putting the values in the above equation; we get:

So;

The molar absorptivity coefficient of erioglaucine is

The absorbance of erioglaucine in distilled water (contaminated with metal ions) is:

The absorbance of distilled water is

So; absorbance of erioglaucine itself is :

Again using Lambert Beer law; we get:

c = 0.198/5.65 M = 0.035 M

The concentration of the erioglaucine is

Explanation:

58750.08068 is the answer i think

personal, she chose to do it.