Question

Mirex(MW = 540) is a fully chlorinated organic pesticide that was manufactured to control fire ants. Due to its structure, mirex is very unreactive; thus, it persists in the environment. Lake Erie water samples have had mirex measured as high as 0.002 μg/L and lake trout samples with 0.002 μg/g. (a) (10 points) In the water samples, what is the aqueous concentration of mirex in units of (i) ppb, (ii) ppt, (iii) μM? (b) (10 points) In the fish samples, what is the concentration of mirex in (i) ppm, (ii) pp

Answer 1

Answer:

A) i) 0.002 ppb (ii) 2ppt (iii) 3.7 x 10^(-6) μM

B) i)0.002 ppm. (ii) 2ppb

Explanation:

A) We know that 1 ppb = 1 μgram per liter, and so the concentration of Mirex in ppb would be 0.002 ppb.

1 ppt = 1 nanogram per liter of water, so the concentration of Mirex in ppt would be 2 ppt;

(0.002 μg/L) (100ng/μg) (1ppt/ng/L) = 2ppt.

Now, MW of Mirex = 540 g/mol ≡ μg/μmol

Thus, 1 μmole = 540 μgram,

Hence, the concentration of Mirex in μmoles would be;

(0.002 μg/L)/(540 μg/μmol) = 3.7 x 10^(-6) μM

B) i) 1 ppm = 1 μgram per gram.

Thus, the concentration of Mirex in ppm would be = 0.002 ppm.

ii) Now, 1 ppb = 1 nanogram per gram.

Thus, concentration of Mirex in ppb would be = (0.002 μg/g) (100ng/μg) (1ppb ng/g) = 2ppb

Answer 2

Answer:

The solution to the given problem is done below.

Explanation:

a)

i) =( 0.002 μg / L )( 1mg / 1000 μg )( 1L / kg )( 1000 mil / 1 billion) = 0.002 ppb

ii) =( 0.002 μg / L )( 1mg / 1000 μg )( 1L / kg )( 1,000,000 mil / 1 trillion) = 2 ppt

iii) =( 0.002 μg / L )( 1 mole / 540g ) = 3.7 x $10^{-6}$μM.

b)

i) =( 0.002 μg / g ) = 0.002 ppm

ii) In solids, ppb = μg/kg

=( 0.002 μg / g )( 1000 mil/ 1 billion) = 2 ppb