A salmon was caught off the west coast of the US with a concentration of compoun

Question

A salmon was caught off the west coast of the US with a concentration of compound "X" in its liver. The amount of "X" was 240 ug in a 75 g liver. The company denied that it did not pollute since it follwed EPA standards and its water emissions were below 10 ppt. The solubility of compound "X" is 5 ppt. The fish was determined to be 1 year old, and passed 100 mL seawater through its gills/respiratory system per minute. The company lost 100 g of compound "X" into the aqueous effluent over one year with an effluent flow of 250,000 L/day. What would the possible accumulation of compound "X" in the fish if it absorbed the entire lost compound from the eater. This means the fish would be in the aqueous environment all its life and there was no oceanic current mixing. Calculate all concentrations and all needed numbers for data for the lawsuit of the company.

Explanation / Answer

Answer.

Lets consider the concentration of compound in the water from the company effluent over one year.

Each day, company pumped 250,000 Litres of effluent into the water body.

Total volume of effluent pumped into the water body in one year = 250,000 x 365 litres

                                                                                            = 9.125 x 107 litres

The company has lost 100 gm of compound "X" in one year.

Hence, the concentration of compound "X" in the water body from the company effluent would be,

                                                  = 100 gm/ 9.125 x 107 litres

                                                  = 10.95 x 10-7 gm/L

                                                  = 1.095 x ug/L

This corresponds to 1.095 ppm which is far above the permitted level of 10 ppt.

The salmon fish passes 100 ml of water through its gills per minute.

In 60 minutes (1 hour) the fish will pass = 100 x 60 ml

In 24 hours (1 day) the fish will pass = 100 x 60 x 24

In 365 days ( 1 year) the fish will pass = 100 x 60 x 24 x 365 = 52560000 ml or 52560 litres

Since the concentration of compound "X" is 1.095 x 10-6 gm /L

The amount of compound "X" present in fish will be = 1.095 x 10-6 gm /L x 52560 L

                                                                          = 57553.2 x 10-6 gm

                                                                          = 575.532 mgs

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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