a) What is the concentration of the dye in a solution with a percent transmittan

Question

a) What is the concentration of the dye in a solution with a percent transmittance reading of 45.8% at 640 nm? b) If the solution in part a was made by taking 20.00 mL of a stock solution and diluting it to 100.00 mL, what is the concentration of the dye in the stock solution?

Concentration (mM)--> Absorbance--> Percent Transmittance (Calculate by me)
0.00 --> 0.000 --> 100

1.00 --> 0.102 --> 79.07

2.00 --> 0.143 --> 71.94

3.00 --> 0.278 --> 52.72

4.00 --> 0.401 --> 39.72

5.00 --> 0.451 --> 35.40

I have found the answer elsewhere, but need to know how to do it myself. Another source came up with 3.53 and when plotted in Excel with a trendline, 3.53 seemed to line right up with 45.8%. Thanks in advance.

Explanation / Answer

You use the equation A = ?lc (Beer-Lambert law)

The wavelength is immaterial, but would normally be selected to be at an absorbance peak for the dye.

You have a mixture of results here. Lets ignore the concentration and molecular weight since you have a molar absorptivity. Lets assume the pathlength is 1 cm. This means

c = A/?l = 1.338/(2.41*10^4 * 1) = 5.552 * 10^-5 M.

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