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. A sample of fish-liver oil has an absorbance (at 329 nm) of .850 units, Suggest a reason for using this wavelength. In what region of the spectrum does this wavelength lie? When 0.1232 g of fish-liver oil is dissolved in 500.0 mL of solvent, the absorbance is 0.724 units. When 1.67 times 10^-3 g of vitamin A is dissolved in 250. mL of solvent, the absorbance is 1.018 units. Calculate the vitamin A concentration in the fish-liver oil. ((You do not have to generate a graph to solve this problem. You can use two data points.))

Explanation / Answer

Vitamin A

= 329 nm

Molecular weight of Vitamin A = 286.45 g/mol

a. The wavelength of 329 nm is used because at this wavelength of there is a maximum absorbance (one of the peaks) for vitamin A.

b. The wavelength of 329 nm is in the region of UV (ultra voilet) of the electromagnetic spectrum

c. Amount = 1.67 x 10-3 g dissolved in 250 mL (0.250L).

The conc of Vitamin A = (1.67 x 10-3 g / 286.45 gmol-1)/0.250 L = 5.83 x 10-6 mol/0.250 L = 2.33x 10-5 M

A = 1.018

c = 2.33 x 10-5 M

l = 1cm (assumed)

A = cl

= A/cl = 1.018 / 2.33x 10-5 M x 1 cm = 4.37 x 104 M-1cm-1

= 4.37 x 104 M-1cm-1

A = cl

A = 0.724

c = A/cl = 0.724 / (4.37 x 104 M-1cm-1 x 1 cm) = 0.165 x 10-4 M

c =  0.165 x 10-4 M = 1.65 x 10-5 mol/L

Molecular weight of Vitamin A = 286.45 g/mol

c = 1.65 x 10-5 mol/L x 286.45 g/mol = 472.64 x 10-5 g / L= 4.72 mg/L

Amount of fish oil taken = 0.1232 g (dissolved in 500 mL)

Amount of Vitamin A in 0.1232 g is 4.72/2 = 2.36 mg

Concentration of Vitamin A in fish oil = 2.36 mg / 0.1232 g = 19 mg/ g

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