(a) determine the dose (in mg/kg-day) for a bioaccumulative chemical with bcf = 103 that is found in water at a concentration of 0.1 mg/l. calculate your dose for a 50 kg adult female who drinks 2 l lake water per day and consumes 30 g fish per day that is caught from the lake. ans. 0.064 mg/kg-d (b) what percent of the total dose is from exposure to the water, and what percent is from exposure to the fish?

0.064 mg/kg/day

6.25% from water, 93.75% from fish

Explanation:

Density of water is 1 kg/L, so the concentration of the chemical in the water is 0.1 mg/kg.

The BCF = 10³, so the concentration of the chemical in the fish is:

10³ = x / (0.1 mg/kg)

x = 100 mg/kg

For 2 L of water and 30 g of fish:

2 kg × 0.1 mg/kg = 0.2 mg

0.030 kg × 100 mg/kg = 3 mg

The total daily intake is 3.2 mg.  Divided by the woman's mass of 50 kg, the dosage is:

(3.2 mg/day) / (50 kg) = 0.064 mg/kg/day

b) The percent from the water is:

0.2 mg / 3.2 mg = 6.25%

And the percent from the fish is:

3 mg / 3.2 mg = 93.75%

answer:

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explanation:

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