3. (30%) (closed system, steady state) Consider a lake having the following prop

Question

3. (30%) (closed system, steady state) Consider a lake having the following properties: Lake volume 3.1 x 10m3 Suspended solids (SS): 1 m3 of SS/ 105 m3 water Biota: 31 m3 SS/water partition coefficient, Kss 104 Biota/water partition coefficient, Kbota-10 The lake contains 1.5 kg of a contaminant. Calculate: (a) Concentration (ug m-3)and % of the total contaminant in the water. (b) Concentration(ug m-3) and % of the total contaminant in the suspended solids. (c) Concentration(ug m") and % of the total contaminant in the biota.

Explanation / Answer

Lake water volume = 3.1 x 107 m3

Volume of SS = 3.1 x 107 / 105 m3 = 3.1 x 102 m3

Volume of Biota = 31 m3

Total mass of contaminant = 1.5 kg

Mass of contaminant in SS / Mass of contaminant in water = 104

Mass of contaminant in Biota / Mass of contaminant in water = 105

Let mass of contaminant in water = m

Total mass of contaminant = m + m 104 + m 105 = 1.5 kg

m = 1.36 x 10-5 kg = 13600 g

a)

Concentration of contaminant in water = m / Lake water volume

= 13600 g / 3.1 x 107 m3

= 4.4 x 10-4 g / m3

% of total contaminant in water = m / 1.5 kg * 100 = 9 x 10-4 %

b)

Concentration of contaminant in SS = m 104 / Volume of SS

= 13600 * 104 g / 3.1 x 102 m3

= 4.4 x 105 g / m3

% of total contaminant in SS = m 104 / 1.5 kg * 100 = 9.09 %

c)

Concentration of contaminant in Biota = m 105 / Volume of Biota

= 13600 * 105 g / 31 m3

= 4.4 x 107 g / m3

% of total contaminant in Biota = m 105 / 1.5 kg * 100 = 90.9 %

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