Chapter 1.4, Problem 1.6BFP

Interpretation Introduction

Interpretation:

The mass of mercury in grams contained in each $\text{mL}$ of lake water is to be calculated. Also, a road map for the solution is to be drawn.

Concept introduction:

Meter square $\left({\text{m}}^{\text{2}}\right)$ is the SI unit of the area. The English equivalent of the area is miles square $\left({\text{mi}}^{\text{2}}\right)$. The conversion factor to convert the area from ${\text{mi}}^{\text{2}}$ to ${\text{ft}}^2$ is,

${\left(0.3048\right)}^2{\text{ m}}^2=1{\text{ ft}}^2$

Mass is a physical quantity and its SI unit is kilograms $\left(\text{kg}\right)$. It is the measure of the quantity of the matter present in an object. The other units of mass are grams, milligrams.

The conversion of one unit into another can be done using a proper conversion factor. Conversion factors are the ratios that relate the two different units of a quantity. It is also known as dimensional analysis or factor label method.

In the unit conversion problems, the given information is multiplied by the conversion factors to obtain the desired information. The unit conversion can be done as follows:

$\left(\cancel{\text{beginning unit}}\right)\left(\frac{\text{Final unit}}{\cancel{\text{beginning unit}}}\right)=\text{Final unit}$

Answer to Problem 1.6BFP

The road map to calculate the mass of mercury in grams contained in each $\text{mL}$ of lake water is as follows:

The mass of mercury in grams contained in each $\text{mL}$ of lake water is $6.0\times {10}^{-7}\text{ g/mL}$.

Explanation of Solution

The road map to calculate the mass of mercury in grams contained in each $\text{mL}$ of lake water is as follows:

The mass of mercury discharged in the lake is $75,000\text{ kg}$.

The conversion factor to convert the mass from kilograms $\left(\text{kg}\right)$ to grams $\left(\text{g}\right)$ is,

$1\text{ kg}=1000\text{ g}$

The mass of mercury in the lake in grams is,

$\begin{array}{c}\text{Mass}\left(\text{g}\right)=75,000\text{ kg}\left(\frac{1000\text{ g}}{1\text{ kg}}\right)\\ =7.5\times {10}^7\text{ g}\end{array}$

The surface area of the lake is $4.5{\text{ mi}}^2$ and the depth of the lake is $35\text{ ft}$.

The conversion factor to convert the area from ${\text{mi}}^{\text{2}}$ to ${\text{ft}}^2$ is,

${\text{1 mi}}^2={\left(5280\right)}^2{\text{ ft}}^2$

The surface area of the lake in ${\text{ft}}^2$ is calculated as:

$\begin{array}{c}\text{Area}\left({\text{ft}}^2\right)=4.5{\text{ mi}}^2\left(\frac{{\left(5280\right)}^2{\text{ ft}}^2}{{\text{1 mi}}^2}\right)\\ =1.25452800\times {10}^8{\text{ ft}}^2\end{array}$

The volume of  the lake is calculated as:

$\text{Volume of lake}\left({\text{ft}}^{\text{3}}\right)=\left(\text{Area of lake}\left({\text{ft}}^2\right)\right)\left(\text{depth of lake}\left(\text{ft}\right)\right)$ (1)

Substitute $1.25452800\times {10}^8{\text{ ft}}^2$ for the area of the lake and $35\text{ ft}$ for depth of lake in equation (1).

$\begin{array}{c}\text{Volume of lake}\left({\text{ft}}^{\text{3}}\right)=\left(1.25452800\times {10}^8{\text{ ft}}^2\right)\left(35\text{ ft}\right)\\ =4.3908\times {10}^9{\text{ ft}}^3\end{array}$

The conversion factor to convert volume from ${\text{ft}}^3$ to ${\text{m}}^{\text{3}}$ is,

$1{\text{ ft}}^3=0.02832\text{}{\text{ m}}^3$

The conversion factor to convert volume from ${\text{m}}^{\text{3}}$ to ${\text{cm}}^{\text{3}}$ is,

$1{\text{ m}}^3=1\times {10}^6{\text{ cm}}^3$

The conversion factor to convert volume from ${\text{cm}}^{\text{3}}$ to $\text{mL}$ is,

${\text{1 cm}}^3=1\text{}\text{ mL}$

The volume of the lake in $\text{mL}$ is,

$\begin{array}{c}\text{Volume of lake}\left(\text{mL}\right)=4.3908\times {10}^9{\text{ ft}}^3\left(\frac{0.02832\text{}{\text{ m}}^3}{1{\text{ ft}}^3}\right)\left(\frac{1\times {10}^6{\text{ cm}}^3}{1{\text{ m}}^3}\right)\left(\frac{1\text{}\text{ mL}}{{\text{1 cm}}^3}\right)\\ =1.24349\times {10}^{14}\text{ mL}\end{array}$

The mass of mercury in grams present in one $\text{mL}$ of lake water is calculated as,

$\text{Mass}\left(\text{g}\right)\text{of mercury per mL}=\frac{\text{Mass of mercury discharged in lake}\left(\text{g}\right)}{\text{Volume of lake}\left(\text{mL}\right)}$ (2)

Substitute $7.5\times {10}^7\text{ g}$ for the mass of mercury discharged in the lake and $1.24349\times {10}^{14}\text{ mL}$ for the volume of the lake in equation (2).

$\begin{array}{c}\text{Mass}\left(\text{g}\right)\text{of mercury per mL}=\frac{7.5\times {10}^7\text{ g}}{1.24349\times {10}^{14}\text{ mL}}\\ =6.0314\times {10}^{-7}\text{ g/mL}\\ =6.0\times {10}^{-7}\text{ g/mL}\end{array}$

Conclusion

The mass of mercury in grams contained in each $\text{mL}$ of lake water is $6.0\times {10}^{-7}\text{ g/mL}$.