Answer
$6.32\cdot 10^5\text{ kg}$
Work Step by Step
To find the total mass of mercury in the lake, we first need to calculate the volume of the lake. We can use the formula: $$\text{volume} = \text{surface area} \times \text{average depth}.$$ First we convert the surface area in square centimeters and the depth in centimeters.
$$\begin{aligned}
1\text{ mi}^2& = 160934.4^2\text{ cm}^2\\
&=2.5899881103\cdot 10^{10}\text{ cm}^2 \\
\text{surface area}&=100\cdot 2.5899881103\cdot 10^{10}\text{ cm}^2\\
&=2.5899881103\cdot 10^{12}\text{ cm}^2\\
1\text{ ft}& = 30.48\text{ cm}\\
20\text{ ft}&=20\cdot 30.48\text{ cm}\\
&=609.6\text{ cm}\\
&=6.096\times 10^2\text{ cm}.
\end{aligned}$$ So, the volume of lake is: $$\begin{aligned}
\text{volume}&=
2.5899881103\cdot 10^{12}\text{ cm}^2\cdot 6.096\cdot 10^2\text{ cm}\\
&=15.7885675204\cdot 10^{14}\text{ cm}^3\\
&=1.57885675204\cdot 10^{15}\text{ cm}^3.
\end{aligned}$$ Now we know that the concentration of mercury in the lake is $0.4\mu\text{g Hg/mL}$ and that $1\text{ mL}=1\text{ cm}^3$, so we can find the total mass of mercury by multiplying the concentration by the volume of the lake: $$\begin{aligned}
&0.4\mu\text{g/mL}\cdot 1.57885675204\cdot 10^{15}\text{ cm}^3\\
&=0.631542700816\cdot 10^{15}\cdot 10^{-9} \text{kg}\\
&\approx 631542\text{ kg}\\
&\approx 6.32\cdot 10^5\text{ kg}.
\end{aligned}$$ The total mass of mercury in the lake is $6.32\cdot 10^5\text{ kg}$.
