Question

) A person weighing 50 kilograms eats 0.6 kg of fish containing 0.5 (ug /gm of fish) mercury (Hg) two times a week. What is the steady state average concentration of Hg in the body of this person over time. Half-life (t_1/2) of Hg in a person is 100 days.

Answer 1

Weight of person = 50 kg

Mass of fist taken per day = 0.6 kg

Amount of mercury in fish = 0.5 g /gm of fish

Half life of Hg in a person = 100 days

Rate of intake of mercury = 600 g (fish) * 0.5 g /gm of fish * 10^-3 mg / g * (2/7) / day

= 0.085 mg / day

Mass of mercury taken in steady state:

= 0.085 mg/day * 100 days

= 8.5 mg

Concentration of mercury taken in the body = 8.5 mg / 50 kg

= 0.17 mg mercury / kg of body

or;

= 0.17 ppm

Hence,

Steady state average concentration of Hg in the body of this person = 0.17 ppm