Gold, which has a density of$\mathbf{19}\mathbf{.}\mathbf{32}\mathbf{}\mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}$, is the most ductile metal and can be pressed into a thin leaf or drawn out into a long fiber. (a) If a sample of gold, with a mass of 27.63 g, is pressed into a leaf of 1.000 µm thickness, what is the area of the leaf? (b) If, instead, the gold is drawn out into a cylindrical fiber of radius 2.500 µm, what is the length of the fiber?

The density of gold, $19.32\mathrm{g}/{\mathrm{cm}}^3$

The mass of gold, $\mathrm{m}=27.63\mathrm{g}$

The thickness of a leaf, role="math" localid="1654513937755" $\mathrm{t}=1.000\mathrm{\mu m}$

The radius of fiber, $\mathrm{r}=2.500\mathrm{\mu m}$

The density of a material (in this case gold) is defined as the mass per unit volume. In this problem, the volume of gold is equal to the volume of gold pressed into a leaf and long fiber.

The expression for density is given as:

$\mathrm{p}=\frac{\mathrm{m}}{\mathrm{v}}$ … (i)

Here, $\mathrm{p}$is the density, $\mathrm{m}$is the mass and $\mathrm{v}$ is the volume.

Using equation (i), the volume of gold is calculated as:

$$\begin{gathered} V=\frac{m}{p} \\ =\frac{27.63\mathrm{g}}{19.32\mathrm{g}/{\mathrm{cm}}^3} \\ =1.430{\mathrm{cm}}^3 \end{gathered}$$

Now, convert the volume ^{$1.430{\mathrm{cm}}^3$}into${\mathrm{m}}^3$.

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

Therefore,

$$\begin{gathered} V=1.430{\mathrm{cm}}^3\times \frac{1\times {10}^{-6}{\mathrm{m}}^3}{1{\mathrm{cm}}^3} \\ =1.430\times {10}^{-6}{\mathrm{m}}^3 \end{gathered}$$

Convert the thickness $1.000\mathrm{\mu m}$into m.

$1.000\mathrm{\mu m}=1\times {10}^{-6}\mathrm{m}$

The expression for the area of the leaf is,

$A=\frac{V}{t}$

Substitute the values in the above expression.

$A=\frac{1.430\times {10}^{-6}{\mathrm{m}}^3}{1\times {10}^{-6}\mathrm{m}}$

Thus, the area of the leaf is $1.430{\mathrm{m}}^2$.

The volume of the cylinder is given as:

$V=A\times L$ … (ii)

Here, A is the cross section area and L is the length.

The cross-section area of cylinder is given as:

$A=\pi r^2$ … (iii)

Here, r is the radius of the cylinder.

From equation (ii) and (iii),

$L=\frac{V}{\pi r^2}$ … (Iv)

Substitute the values ofr andV in equation (iv).

$$\begin{gathered} L=\frac{1.430\times {10}^{-6}{\mathrm{m}}^3}{3.142\times {(2.500\times {10}^{-6}\mathrm{m})}^2} \\ =7.284\times {10}^4\mathrm{m} \end{gathered}$$