A).The bioconcentration factor (BCF) of DDT is often expressed in a logarithmic

Question

A).The bioconcentration factor (BCF) of DDT is often expressed in a logarithmic form, so log10 BCF = 5.

A river was sampled and its water was found to have a DDT concentration of 0.025 ppm. Based on this, what is the concentration of DDT expected to be found in the fat of an organism living in the river?

B).The LD50 for DDT is quoted as 500 mg kg-1

(i) State what is meant by the term LD50, explaining what the     units associated with it mean.(2marks)

(ii)The FDA (Food and Drug Administration in the USA) sets a limit of 5 ppm (5 mg kg-1) for the DDT content of fish. What mass of fish would an adult weighing 80 kg need to consume to reach their LD50, assuming the fish was contaminated to this limit? ...(6 marks)

Explanation / Answer

A).The bioconcentration factor (BCF) of DDT is often expressed in a logarithmic form, so log10 BCF = 5.

A river was sampled and its water was found to have a DDT concentration of 0.025 ppm. Based on this, what is the concentration of DDT expected to be found in the fat of an organism living in the river?

Solution :-

Log BCF = 5

Therefore BCF = antilog [5]

                          = 1*10^6

Therefore

Lets calculate the concentration of the DDT in living organism

0.025 ppm * 1*10^6 = 2500 ppm

So the concentration of the DDT is 2500 ppm

That is 2500 mg / kg

B).The LD50 for DDT is quoted as 500 mg kg-1

(i) State what is meant by the term LD50, explaining what the     units associated with it mean.(2marks)

Solution :- LD50 is the lithal dose of the contaminant by which 50 % of the population dies which is under study.

Unit associated with it means pertuclar amount of the contaminant per kg body weight is lithal dose.

(ii)The FDA (Food and Drug Administration in the USA) sets a limit of 5 ppm (5 mg kg-1) for the DDT content of fish. What mass of fish would an adult weighing 80 kg need to consume to reach their LD50, assuming the fish was contaminated to this limit?

Solution :-

5 ppm = 5 mg / kg

So 80 kg * 5 mg / 1 kg = 400 mg

Now lets calculate the mass of the fish needed

400 mg / 2500 mg/ kg = 0.160 kg fish

So it need 0.160 kg fish to be eaten to reach the limit.

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