Answer to Question #131432 in Chemistry for badria alrubah

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.
Kuwait Bay water samples have had mirex measured
as high as 0.002 µg/L and Zubaidi fish samples with 0.002 µg/g.
(a) In the water samples, what is the aqueous concentration of mirex in units of
(i) Ppbm
(ii) Pptm
(iii) µM
(b) In the fish samples, what is the concentration of mirex in
(i) Ppmm
(ii) ppbm

Answer 1

(a)

(i) ppbm - parts per billion. 1 ppbm = 1 µg of substance per L of solution (µg/L). From here:

0.002 µg/L = 0.002 ppbm = 2 × 10^-3 ppbm

(ii) ppt - parts per trillion. 1 ppt = 1 ng of substance per L of solution (ng/L). From here:

0.002 µg/L = 0.002 × 10³ ng/L = 2 ng/L = 2 pptm

(ii) M = n / V = m / MrV

where M - molar concentration, n - number of moles, V - volume, Mr - molecular weight. From here:

M(Mirex) = m / MrV /= 0.002 µg / (540 g/mol × 1 L) = 0.002 × 10^-6 g / (540 g/mol × 1 L) = 3.7 × 10^-12 mol/L = 3.7 × 10^-6 µmol/L = 3.7 × 10^-6 µM

(b)

(i) As 1 pptm = 1 ng of substance per kg (ng/kg):

0.002 µg/g = 0.002 mg/kg = 0.002 × 10⁶ ng/kg = 2 × 10³ ng/kg = 2 × 10³ pptm

(ii) As 1 ppbm = 1 µg of substance per kg (µg/kg):

0.002 µg/g = 0.002 mg/kg = 0.002 × 10³ µg/kg = 2 µg/kg = 2 ppbm