Question

 

You wish to construct an absorbance versus concentration (A vs. C) plot for a FD&C food dye. Suppose you have a stock solution that is 2.13 x 10^-5 M in this dye. Complete the following table (fill in the blanks) showing how the different calculations were performed

Stock Concentration	Volume of Stock (mL)	Volume of Diluent (mL)	Diluted Sample Concentration (M)
2.13 x 10-5 M	5.00	0.00
	3.67		1.56 x 10-5
	2.50	2.50	1.07 x 10-5
	1.33	3.67
	0.33		1.41 x 10-6

You wish to determine the molar absorptivity ε for another food dye

Using only the data below, find ε for this dye by constructing an absorbance versus concentration plot. Properly scale the plot if necessary. The slope of this plot is ε. Include the R² value and the equation of the linear trend-line (both from Excel). You will likely need to adjust the sig figs for the equation of the linear trend-line and R² in Excel. Properly title the plot and label the axes correctly. Paste-in the plot used to find ε. State your ε value (the slope). Use correct sig figs and units.

Absorbance	Concentration (M)
0.842	5.94 x 10-6
0.641	4.36 x 10-6
0.388	2.97 x 10-6
0.202	1.58 x 10-6
0.033	3.92 x 10-7

Answer 1

moles of stock solution in 5ml of 2.13*10-5M= molarity* volume in L= 2.13*10-5*5/1000 moles

since no dilution is done, the concentration remains at 2.13*10-5M

when 3.67ml is added to dilute the concentration to 1.56*10-5,

from C1V1=C2V2

V2= C1V1/C2= 2.13*10-5*3.67/(1.56*10-5) =5ml, volume to be added= 5-3.67= 1.33 ml

when 2.5ml is added to 2.5 ml of stock, the concentration =2.13*10-5*2.5/5 = 1.07*10-5M

when 1.33 ml of stock is diluted by addiing another 3.67ml, the concentration is 2.13*10^-5*1.33/5=5.7*10-6M

when 0.33ml of 2.13*10-5 is diluted to 1.41*10-6M

2.13*10^-5*0.33= x*1.41*10-6, x= final volume = 5ml, volume to be additonally added= 5-0.33= 4.67ml

since A= ebC, a plot of concentration vs C is created and is shown below

from the plot R2 value = 0.993 and e=14112/M