Question

Question 2 Compound "Y" (molar mass = 480.5 g moll) used as a food coloring. Following experiment was carried out to find the concentration of "Y" in a beverage. A stock solution of compound "Y" (concentration = 7.50 0 10-4 mol L-) was used to prepare a series of standard solutions. Calibrated pipets and 100.0 mL volumetric flasks were used for these preparations. Then, absorbance of these solutions were measured at 460 nm using 1-cm cuvettes. A dilution was carried out for the beverage (5:100 dilution - using 5 mL calibrated pipet and a 100.0 mL volumetric flask). Absorbance of the diluted beverage sample was also recorded at 460 nm using 1-cm cuvette. Table 1: Calibrated volume of pipets Calibrated volume of the 1 mL pipet Calibrated volume of the 2 mL pipet Calibrated volume of the 5 mL pipet Select a value within the given range 1.001 - 1.006 mL 1.998 - 2.003 mL 4.98 - 5.01 mL Table 2: Absorbance measurements Standard solution 1 prepared with 1 mL pipet Standard solution 2 prepared with 2 mL pipet Standard solution 3 prepared with 5 mL pipet Diluted beverage sample Absorbance at 460 nm 0.132 0.276 0.675 0.355 Part 1 Calculate the concentrations of the three standard solutions (use table 1 to select a calibrated volume for each pipet, you can select any value within the given range). Part 2 Now calculate the molar absorptivity of "Y" (at 460 nm) for each standard solution using the corresponding absorbance values. Then, calculate the average molar absorptivity. Part 3 Now calculate the concentration of "Y" in the diluted beverage sample using the average molar absorptivity from part 2. Then, calculate the concentration of "Y" in the original beverage in ppm (parts per million).

Answer 1

Std #	Vol of Stock Solution, mL	Total volume, mL	Conc, M	Abs	ε
1	1	100	0.0000075	0.132	17600
2	2	100	0.000015	0.276	18400
3	5	100	0.0000375	0.675	18000
Unkn		100		0.355	avg= 18000

Given that -

Conc of stock = 7.5 * 10^-4 M

Part 1-

1 ml of this is stock is poured in 100 ml of water then new conc =

M1V1 = M2V2

7.5 * 10^-4 M * 1 = M2 * 100

M2 =  7.5 * 10^-4 M / 100

M2 = 0.0000075 M For Std 1

Likewise

M2 = (7.5 * 10^-4 * 2 ) / 100

M2 = 0.000015 M for std 2

M2 = (7.5 * 10^-4 * 5 ) / 1000

M2 = 0.0000375 M For std 3

Now, Plotting a calibration curve between Abs vs Conc,

Calibration curve B (y-intercept) = 0.000692307692307663 +/- 0.00695186322987325 A (slope) = 18,015.3846153846 +/- 293.116290511656 Abs 0.1 52-06 1e-05 1.5e-05 2e-05 2.5e-05 3e-05 3.5e-05 4e-05 Conc,M

we got -

B (y-intercept) = 0.000692307692307663 +/- 0.00695186322987325

A (slope) = 18,015.3846153846 +/- 293.116290511656

Thus on comparing with beer-lambert equation, A = ε C l

Part 2

we got slope = ε = 18,015.3846 = molar absorbtivity

And Average ε from calculation = 18000 M-1 cm-1

Part 3-

Thus diluted concentration of Y =

0.355 = 18000 * 1 * C

C = 0.355 / 18000

C = 0.0000197 M