Question

Mercury poisoning is a debilitating disease that is often fatal. In the human body, mercury reacts with essential enzymes leading to irreversible inactivity of these enzymes. If the amount of mercury in a polluted lake is 0.4 ug Hg/ml, what is the total mass in kilograms of the mercury in the lake? (The lake has an area of 100 mi^2 and an average depth of 20 ft.)

Answer 1

Solution :-

converting area to m2

100 mi2 * 2.59*10^6 m2 / 1 mi2 = 2.59*10^8 m2

converting depth to m

20 ft = 6.096 m

Lets first calculate the volume of the lake

Volume of lake = area * depth

                            = 2.59*10^8 m2 * 6.096 m

                           = 1.58*10^9 m3

1.58*10^9 m3 * 1*10^6 mL / 1 m3 = 1.58*10^15 ml

Now lets calculate the amount of the Hg in this volume

1.58*10^15 ml * 0.4 ug / ml = 6.32*10^14 ug Hg

Now lets convert the ug of Hg to kg

6.32*10^14 ug * 1 kg / 1*10^9 ug = 6.32*10^5 kg Hg

So the amount of the Hg present is 6.32*10^5 kg Hg