# Question 539ce

Jun 9, 2017

$8 \cdot {10}^{5}$ $\text{kg}$

#### Explanation:

Start by converting the surface area of the lake to square meters. This will then allow you to calculate the volume of the lake in cubic meters.

Use the fact that

$\text{1 km"^2 = "1 km" * "1 km}$

to get

100 color(red)(cancel(color(black)("km"))) * color(red)(cancel(color(black)("km"))) * (10^3color(white)(.)"m")/(1color(red)(cancel(color(black)("km")))) * (10^3color(white)(.)"m")/(1color(red)(cancel(color(black)("km")))) = 1 * 10^8 ${\text{m}}^{2}$

Now, you can calculate the volume of the lake by multiplying its surface area by its average depth.

In your case, you will have

$1 \cdot {10}^{8} \textcolor{w h i t e}{.} \text{m"^2 * "20 m} = 2 \cdot {10}^{9}$ ${\text{m}}^{3}$

Now, in order to be able to use the known concentration of arsenic, you must convert the volume to cubic decimeters.

2 * 10^9 color(red)(cancel(color(black)("m"))) * color(red)(cancel(color(black)("m"))) * color(red)(cancel(color(black)("m"))) * "10 dm"/(1color(red)(cancel(color(black)("m")))) * "10 dm"/(1color(red)(cancel(color(black)("m")))) * "10 dm"/(1color(red)(cancel(color(black)("m")))) = 2 * 10^12 ${\text{dm}}^{3}$

Next, use the fact that

$\text{1 dm"^3 = "1 L}$

and

$\text{1 L} = {10}^{3}$ $\text{mL}$

to convert the volume to milliliters

2 * 10^(12) color(red)(cancel(color(black)("dm"^3))) * (1color(red)(cancel(color(black)("L"))))/(1color(red)(cancel(color(black)("dm"^3)))) * (10^3color(white)(.)"mL")/(1color(red)(cancel(color(black)("L")))) = 2 * 10^(15) $\text{mL}$

Next, calculate the mass of arsenic in micrograms

2 * 10^15 color(red)(cancel(color(black)("mL"))) * (0.4color(white)(.)mu"g")/(1color(red)(cancel(color(black)("mL")))) = 8 * 10^(14)# $\mu \text{g}$

Finally, use the fact that

$\text{1 g} = {10}^{6}$ $\mu \text{g}$

and

$\text{1 kg} = {10}^{3}$ $\text{g}$

to convert the mass to kilograms

$8 \cdot {10}^{14} \textcolor{red}{\cancel{\textcolor{b l a c k}{\mu \text{g"))) * (1color(red)(cancel(color(black)("g"))))/(10^6color(red)(cancel(color(black)(mu"g")))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)(8 * 10^5color(white)(.)"kg}}}}$

The answer is rounded to one significant figure.