Fish liver oil is a good source of vitamin A, whose concentration is measured sp

Question

Fish liver oil is a good source of vitamin A, whose concentration is measured spectrometrically at a wavelength of 329 nm. Calibration with pure vitamin A yields the absorbance of 1.018 when 1.67x10-3 g of vitamin A is dissolved in 250.0 mL of solvent. When 0.1240 g sample of a fish-liver oil was dissolved in 500.0 mL of solvent, the absorbance was 0.715. What was the mass % of vitamin A in the sample of this fish-liver oil? (Assume that Beer's Law is valid in the range of vitamin A concentrations given and that the volume of the solven does not change significantly after dissolution of vitamin A.)

Explanation / Answer

From Beer's law,

Absorbance A = ecl

A = kc   where k= constant

Given that vitamin A yields the absorbance of 1.018 when 1.67x10-3 g of vitamin A is dissolved in 250.0 mL of solvent.

Then,
1.018 = k x 0.00167

k = 609.6

For the unknown, assuming a straight line relationship, we want to compare the same volume of material;

Therefore, if 0.124 g of fish liver oil dissolved it in 500 mL volume, then A would be twice that in the problem or 0.715*2.

0.715*2 = k x c

0.715 x 2 = 609.6 x c

C = 0.0023 grams/mL

Hence,

% vitamin A = (mass of vitamin A in the sample/0.124 g)*100

                  = (0.0023 /0.124 g)*100

                  = 1.85 %

Therefore,

mass % of vitamin A in the sample of fish-liver oil = 1.85 %

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.