Question

3. A fish captured from a mining-contaminated lake was dissected and its muscle and liver samples were taken, freeze dried, and ground to powder. The water content in the muscle and liver sample is 84.0% and 71.0% respectively. 0.2500g of a dried sample powder was digested with a strong oxidant reagent and acid under a high temperature. The digest (aqueous solution of a completely digested sample) was transferred into a 25.00mL volumetric flask. The digest was diluted 10 and 25 times for muscle and liver respectively. The concentration of Se measured in the diluted sample solution was 10.5µg/L and 8.3µg/L for the muscle and liver sample. Calculate how much Se is in the muscle and liver samples, express them in both mg/kg dry and wet sample weight. If the concentration of Se in lake water is 0.7µg/L, how much Se is accumulated in muscle and liver? What are the results.

Answer 1

ANSWER

MUSCLE TISSUE:

0.2500 g of dry sample of fish is analysed and 10.5µg/L and 8.3µg/L of Se for muscle and liver tissue was found respectively.

It implies 0.2500 g dry weight of muscle contain  10.5µg of Se

10.5µg = 10.5 X 10^-3 mg

Because 1 µg = 10^-3 mg

Thus 0.2500 g of muscle contain 10.5 X 10^-3 mg of Se

Equating

0.2500 g = 10.5 X 10^-3 mg of Se

1 g = 10.5 X 10^-3 / 0.2500 = 42.0 X 10^-3 mg

It implies 1 g of dry weight of muscle contains 42.0 X 10^-3 mg of Se

Therefore 1Kg or 1000g of sample will contain 42.0 X 10^-3 X 1000 = 42.0 mg of Se

Thus the concentration of Se in Muscle is 42.0 mg / Kg dry mass.

Concentration in terms of wet mass:

Muscle tissue is 84% water , In other words it is 100 - 84 = 14% dry mass

Equating

Dry mass = 14 % total muscular mass

Or dry mass = 14 / 100 X Muscular mass

Muscular mass = 100 / 14 X dry mass

Given 0.2500g of dry mass

Muscular mass = 100 / 14 X 0.2500g = 1.785g

It means 0.2500 g of dry mass is = 1.785 g of muscular tissue

Hence we can say 1.785 g of Muscle contain 10.5 X 10^-3 mg of Se

because 0.2500g of dry mass = 10.5 X 10^-3 mg of Se

Concentration of Se in muscle tissue = (Mass of Se / Mass of Muscle ) X 1000

(10.5 X 10^-3 mg of Se / 1.785 ) X 1000g

= 5.88 mg/Kg of muscle

Concentration of Se in Lake = 0.7 mg / L = 0.7 mg/Kg

Taking the density of lake water as 1g/mL

Concentration of Se in Lake = 0.7 mg / L = 0.7 mg/Kg

Thus accumulated Se = 5.88 mg - 0.7 mg = 5.18 mg / Kg

LIVER TISSUE:

8.3µg/L = 8.3 X 10^-3 mg/L

Thus there are 8.3 X 10^-3 mg of Se in 0.2500 g of dry mass of liver

Concentration = (8.3 X 10^-3 mg / 0.2500) X 1000 = 33.2 mg / Kg dry mass

Liver mass = Dry mass X 100 / 29 (shown above)

Liver mass = 0.2500g X 100 / 29 = 0.86 g

Concentration = (8.3 X 10^-3 mg / 0.86 ) X 1000 = 9.65 mg / Kg of liver tissue.

Accumulated Se = 9.65 mg - 0.7 mg = 8.95 mg/Kg