Chapter 1, Problem 1.38P

a)

Interpretation Introduction

Interpretation:Area of gold foil in ${\text{in}}^{\text{2}}$ that has thickness of $1.6\times {10}^5\text{ in}$ 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 1.38P

Area of gold foil is $11908.9{\text{ in}}^2$ .

Explanation of Solution

Since one troy ounce is equivalent to 31.1 grams, mass of gold can be calculated as follows:

  $\begin{array}{c}\text{Mass of gold}=\left(2\text{ troy oz}\right)\left(\frac{31.1\text{ g}}{1\text{ troy oz}}\right)\\ =62.2\text{ g}\end{array}$

The expression to calculate density of goldis as follows:

  $d=\frac{M}{V}$

Where,

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

Rearrange above equation for $V$ .

  $V=\frac{M}{d}$

The value of $M$ is $62.2\text{ g}$ .

The value of $d$ is $19.3{\text{ g/cm}}^3$ .

Substitute the values in above equation.

  $\begin{array}{c}V=\frac{M}{d}\\ =\frac{\text{62}\text{.2 g}}{19.3{\text{ g/cm}}^3}\\ =3.22{\text{ cm}}^3\end{array}$

The formula to calculate volume of gold is as follows:

  $V=At$

Where,

• $V$is volume of gold.
• $A$is area of gold foil.
• $t$is thickness of gold foil.

Rearrange above equation for $A$ .

  $A=\frac{V}{t}$

The value of $V$ is $3.22{\text{ cm}}^3$ .

The value of $t$ is $1.6\times {10}^{-5}\text{ in}$ .

Substitute the values in above equation.

  $\begin{array}{c}A=\frac{V}{t}\\ =\frac{\left(3.22{\text{ cm}}^3\right){\left(\frac{1\text{ in}}{2.54\text{ cm}}\right)}^3}{1.65\times {10}^{-5}\text{ in}}\\ =11908.9{\text{ in}}^2\end{array}$

Hence, area of gold foil is $11908.9{\text{ in}}^2$ .

b)

Interpretation Introduction

Interpretation: Area of gold foil in ${\text{cm}}^{\text{2}}$ that can be made from $\$75.00$ worth of goldis 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 1.38P

Area of gold foil that can be made from $\$75.00$ worth of goldis $144190{\text{ cm}}^2$ .

Explanation of Solution

Since $\$20.00$ can make one troy ounce of gold, mass of gold that can be made from $\$75.00$ can be calculated as follows:

  $\begin{array}{c}\text{Mass of gold}=\left(\$75.00\right)\left(\frac{1\text{ troy oz}}{\$20.00}\right)\left(\frac{31.1\text{ g}}{1\text{ troy oz}}\right)\\ =116.625\text{ g}\end{array}$

The expression to calculate density of gold is as follows:

  $d=\frac{M}{V}$

Where,

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

Rearrange above equation for $V$ .

  $V=\frac{M}{d}$

The value of $M$ is $\text{116}\text{.625 g}$ .

The value of $d$ is $19.3{\text{ g/cm}}^3$ .

Substitute the values in above equation.

  $\begin{array}{c}V=\frac{M}{d}\\ =\frac{\text{116}\text{.625 g}}{19.3{\text{ g/cm}}^3}\\ =6.043{\text{ cm}}^3\end{array}$

The formula to calculate volume of gold is as follows:

  $V=At$

Where,

• $V$is volume of gold.
• $A$is area of gold foil.
• $t$is thickness of gold foil.

Rearrange above equation for $A$ .

  $A=\frac{V}{t}$

The value of $V$ is $6.043{\text{ cm}}^3$ .

The value of $t$ is $1.6\times {10}^{-5}\text{ in}$ .

Substitute the values in above equation.

  $\begin{array}{c}A=\frac{V}{t}\\ =\frac{6.043{\text{ cm}}^3}{\left(1.65\times {10}^{-5}\text{ in}\right)\left(\frac{2.54\text{ cm}}{1\text{ in}}\right)}\\ =144190{\text{ cm}}^2\end{array}$

Hence, $144190{\text{ cm}}^2$ of gold foil can be made from $\$75.00$ worth of gold.