Question

if the concentration of mercury in the water of a polluted lake is 0.200 ug (micrograms) per liter of water, what is the total mass of mercury in the lake, in kilograms, if the lake has a surface area of 13.5 square miles and an average depth of 28.0 feet

Answer 1

given that;

concentration of mercury in the water of a polluted lake is 0.200 ug (micrograms) per liter of water,

step I:

first convert the unit of surface area from square miles to square feet

we know that

1.0 square miles = 2.788 *10^ 7 square feet

Then ;

13.5 square miles * 2.788 *10^ 7 square feet /1.0 square miles

= 3.76*10^8 square feet


Step II: Now calculate volume of lake water in cubic feet

Volume = surface area * depth

= 3.76*10^8 square feet *28.0 feet

= 1.0528 x 10^10 ft^3

step III:

Then change the unit of volume of lake from cubic feet to liter as follows:

1.0528 x 10^10 ft^3 * (7.48 gallons/ ft^3) * (3.785 liters/gallon)

= 2.98 x 10^11 liters

Step IV: now calculate ethe amount of Hg in micro grams as follows:

0.200 ug (micrograms) / liter of water *2.98 x 10^11 liters

= 5.96*10^10 ug (micrograms)

Step V: lastly Change the unit of concentration of mercury to kg/liter

(5.96*10^10 ug (micrograms)*(1 g/1 x 10^6 microgram)*(1 kg/1 x 10^3 g)

= 59.61 kg Hg


Answer = 59.61 kg mercury