Chapter 1, Problem 1.71P

a)

Interpretation Introduction

Interpretation:Grams of gold that are present in oceans is to be determined.

Concept introduction:Density is defined as mass per unit volume. It is denoted by $d$ . The expression to calculate density is as follows:

  $d=\frac{M}{V}$

Where,

• $d$is the density.
• $M$is mass.
• $V$is volume.

Answer to Problem 1.71P

  $8.0\times {10}^{12}\text{ g}$ ofgold is present in ocean.

Explanation of Solution

The formula to calculate volume of ocean is as follows:

  $V=Ah$

Where,

• $V$isvolume of ocean.
• $A$issurface area of ocean.
• $h$is depth of ocean.

The value of $A$ is $3.63\times {10}^8{\text{ km}}^2$ .

The value of $h$ is $3800\text{ m}$ .

Substitute the values in above equation.

  $\begin{array}{c}V=Ah\\ =\left(3.63\times {10}^8{\text{ km}}^2\right)\left(\frac{{10}^6{\text{ m}}^2}{1{\text{ km}}^2}\right)\left(3800\text{ m}\right)\\ =1.3794\times {10}^{18}{\text{ m}}^3\end{array}$

The expression to calculate density of ocean is as follows:

  $d=\frac{M}{V}$

Where,

• $d$is the density of ocean.
• $M$is mass of ocean.
• $V$is volume of ocean.

Rearrange above equation for $M$ .

  $M=dV$

The value of $d$ is $5.8\times {10}^{-9}\text{ g/L}$ .

The value of $V$ is $1.3794\times {10}^{18}{\text{ m}}^3$ .

Substitute the value in above equation.

  $\begin{array}{c}M=dV\\ =\left(5.8\times {10}^{-9}\text{ g/L}\right)\left(1.3794\times {10}^{18}{\text{ m}}^3\right)\left(\frac{{10}^3\text{ L}}{1{\text{ m}}^3}\right)\\ =8.00052\times {10}^{12}\text{ g}\end{array}$

Hence, $8.0\times {10}^{12}\text{ g}$ of gold is present in ocean.

b)

Interpretation Introduction

Interpretation:Cubic meters of gold that are present in oceans are to be determined.

Concept introduction:Intensive properties depend on nature of substance but extensive properties depend on amount of substance. Intensive properties are same for specific substance while extensive properties are additive in nature. Volume is an example of extensive property.

Volume is defined as amount of space occupied by substance and it is utilized to measure size in three dimensions.

Answer to Problem 1.71P

  $4.1\times {10}^5{\text{m}}^3$ ofgold is present in ocean.

Explanation of Solution

Formula to calculate volume of gold as follows:

  $\text{Volume }\left(V\right)\text{ of gold}=\frac{\text{Mass }\left(M\right)\text{of gold}}{\text{Density }\left(\rho \right)\text{of gold}}$

Mass of goldis $8.00052\times {10}^{12}\text{g}$ .

Density of gold is $\text{19}\text{.3}{\text{g/cm}}^3$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{Volume of gold}=\left(8.00052\times {10}^{12}\text{g}\right)\left(\frac{1{\text{cm}}^3}{19.3\text{g}}\right)\left(\frac{{\left(0.01\text{m}\right)}^3}{{\left(1\text{cm}\right)}^3}\right)\\ =4.14535\times {10}^5{\text{m}}^3\\ =4.1\times {10}^5{\text{m}}^3\end{array}$

c)

Interpretation Introduction

Interpretation:Value of gold present in oceans is to be determined.

Concept introduction:Conversion factor is ratio that is widely used to describe relation between two different units. In order to convert one unit into another, original unit is multiplied with proper conversion factor to yield new unit. This is done as mentioned below.

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

Answer to Problem 1.71P

The value of gold present in ocean is $\$95.18\times {10}^{12}$ .

Explanation of Solution

The formula to calculate volume of ocean is as follows:

  $V=Ah$

Where,

• $V$isvolume of ocean.
• $A$issurface area of ocean.
• $h$is depth of ocean.

The value of $A$ is $3.63\times {10}^8{\text{ km}}^2$ .

The value of $h$ is $3800\text{ m}$ .

Substitute the values in above equation.

  $\begin{array}{c}V=Ah\\ =\left(3.63\times {10}^8{\text{ km}}^2\right)\left(\frac{{10}^6{\text{ m}}^2}{1{\text{ km}}^2}\right)\left(3800\text{ m}\right)\\ =1.3794\times {10}^{18}{\text{ m}}^3\end{array}$

The expression to calculate density of ocean is as follows:

  $d=\frac{M}{V}$

Where,

• $d$is the density of ocean.
• $M$is mass of ocean.
• $V$is volume of ocean.

Rearrange above equation for $M$ .

  $M=dV$

The value of $d$ is $5.8\times {10}^{-9}\text{ g/L}$ .

The value of $V$ is $1.3794\times {10}^{18}{\text{ m}}^3$ .

Substitute the value in above equation.

  $\begin{array}{c}M=dV\\ =\left(5.8\times {10}^{-9}\text{ g/L}\right)\left(1.3794\times {10}^{18}{\text{ m}}^3\right)\left(\frac{{10}^3\text{ L}}{1{\text{ m}}^3}\right)\\ =8.00052\times {10}^{12}\text{ g}\end{array}$

Since $1\text{ troy oz}$ is equivalent to $31.1\text{ g}$ , mass of gold can be calculated as follows:

  $\begin{array}{c}M=\left(8.00052\times {10}^{12}\text{ g}\right)\left(\frac{1\text{ troy oz}}{31.1\text{ g}}\right)\\ =\text{0}\text{.25725}\times {10}^{12}\text{ troy oz}\end{array}$

Since value of $1\text{ troy oz}$ gold is $\$370.00$ , value of gold can be calculated as follows:

  $\begin{array}{c}\text{Value of gold}=\left(\text{0}\text{.25725}\times {10}^{12}\text{ troy oz}\right)\left(\frac{\$370.00}{\text{1 troy oz}}\right)\\ =\$95.18\times {10}^{12}\end{array}$

Hence, value of gold present in ocean is $\$95.18\times {10}^{12}$ .