Question

If the concentration of mercury in the water of a polluted lake is 0.33 µg/L of water, what is the total mass of mercury in the lake, in kilograms, if the lake has a surface area of 10.7 mi2 and an average depth of 45 feet? (Hint: get the volume of the lake first.)

Answer 1

We start by finding out how many liters of water are in the lake.

Then,we change 0.33 microgram/L to kilogram/L

Finally, multiply the concentration of mercury in kg/L with the volume of lake water in liter.

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So,

1. convert 10.7 square miles to square feet

10.7 mi^2 * (640 acre/ mi^2) * (43560 ft^2/acre)

= 298298880 ft^2

2. Now calculate volume of lake water in cubic feet

298298880 ft^2 * 45 ft

= 13423449600 ft^3

3. Now change the volume of lake water to liter

13423449600 ft^3 * (7.48 gallons/ ft^3) * (3.785 liters/gallon)

= 380042020385.28 liters

4. Now Change concentration of mercury to kg/liter

(0.33 microgram/liter)*(1 g/1 x 10^6 microgram)*(1 kg/1 x 10^3 g)

= 0.00000000033 kg/liter

5. Finally Find the mass of mercury

(0.00000000033kg/liter) * (380042020385.28) liters

= 125.41 kg mercury