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:

output?

explanation:

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%