Chapter 1, Problem 85AP

Interpretation Introduction

Interpretation:

The total mass of sodium chloride in the given sample of sea water in kilograms and in tons is to be determined.

Concept introduction:

In order to convert from one unit to another, there are certain relationships between those units. These relationships are called conversion factors.

Dimensional analysis is used to set up and solve a unit conversion problem using conversion factors.

The appropriate conversion factor for any equality is selected in such a way that it results in the proper unit cancellation.

Answer to Problem 85AP

Solution: $4.8\times {10}^{19}\text{ kg}$ and $5.3\times {10}^{16}\text{ tons}$

Explanation of Solution

Given information:

The volume of seawater is $\text{ 1}\text{.5}\times {10}^{21}\text{ L}$.

The seawater contains $\text{3}\text{.1\%}$ sodium chloride by mass.

$1\text{ ton}=2000\text{ lb}$

$1\text{ lb}=453.6\text{ g}$.

The total amount of sodium chloride present in $\text{1}\text{.5}\times {10}^{21}\text{ L}$ seawater can be evaluated as follows:

$\begin{array}{c}\text{Density=}\frac{\text{mass}}{\text{volume}}\\ \left(\frac{\text{1}\text{.03g}}{\text{1 mL}}\right)\text{=}\frac{\text{Mass of NaCl}}{\text{1}\text{.5}\times {\text{10}}^{\text{21}}\text{L}}\\ \left(\frac{\text{1}\text{.03g}}{\text{1 mL}}\right)\text{=}\frac{\text{Mass of NaCl}}{\text{1}\text{.5}\times {\text{10}}^{\text{21}}\text{L}}\end{array}$

$\begin{array}{c}\text{Mass of NaCl=}\left(\frac{\text{1}\text{.03g}}{\text{1 mL}}\right)\times \text{(1}\text{.5}\times {\text{10}}^{\text{21}}\text{ L}\times \left(\frac{\text{1000 mL}}{\text{1 L}}\right)\\ \text{3}\text{.1\% mass of NaCl=}\left(\frac{\text{1}\text{.03g}}{\text{1 mL}}\right)\times \left(\text{1}\text{.5}\times {\text{10}}^{\text{21}}\text{ L}\right)\times \left(\frac{\text{1000 mL}}{\text{1 L}}\right)\times \left(\frac{\text{3}\text{.1 g NaCl}}{\text{100 g}}\right)\end{array}$

$\begin{array}{c}\text{Amount of NaCl}=\left(\frac{\text{1}\text{.03 }\cancel{\text{g}}}{1.0\cancel{\text{mL}}}\right)\left(1.5\times {10}^{21}\cancel{\text{L}}\right)\left(\frac{1000\cancel{\text{mL}}}{1\cancel{\text{L}}}\right)\left(\frac{\text{3}\text{.1 gNaCl}}{100\cancel{\text{g}}}\right)\\ =4.8\times {10}^{22}\text{ g}\end{array}$

Hence, the total amount of sodium chloride present in the seawater is $4.8\times {10}^{22}\text{ g}$.

Convert grams into kilograms by using the following conversion factor:

$\begin{array}{c}\text{Mass of NaCl}=\left(4.8\times {10}^{22}\cancel{\text{g}}\right)\left(\frac{1\text{ kg}}{1000\cancel{\text{g}}}\right)\\ =4.8\times {10}^{19}\text{ kg}\end{array}$

Hence, the mass of sodium chloride in kilograms is $4.8\times {10}^{19}\text{ kg}$.

One pound is equivalent to $453.6\text{ g}$.

One ton is equivalent to $\text{2000 lb}$.

Convert the mass of sodium chloride in tons as follows:

$\begin{array}{c}\text{Mass of NaCl}=\left(4.8\times {10}^{22}\cancel{\text{g}}\right)\left(\frac{1\cancel{\text{lb}}}{\text{453}\text{.6 }\cancel{\text{g}}}\right)\left(\frac{1\text{ ton}}{2000\cancel{\text{lb}}}\right)\\ =5.3\times {10}^{16}\text{ tons}\end{array}$

Conclusion

The total mass of sodium chloride in kilograms and in tonsin the given sample of sea water is calculated to be $4.8\times {10}^{19}\text{ kg}$ and $5.3\times {10}^{16}\text{ tons}$, respectively.