Question

A ${\rm{25}}{\rm{.0 \,mm }} \times {\rm{ 40}}{\rm{.0 \,mm}}$ piece of gold foil is ${\rm{0}}{\rm{.25 \,mm}}$ thick, the density of gold is ${\rm{19}}{\rm{.32 \,g/c}}{{\rm{m}}^{\rm{3}}}$ . How many gold atoms are in the sheet? (Atomic wt ${\rm{Au = 197}}{\rm{.0}}$)
A. $7.7{\rm{ }} \times {\rm{ }}{10^{23}}$
B. $1.5{\rm{ }} \times {\rm{ }}{10^{23}}$
C. ${\rm{4}}{\rm{.3 }} \times {\rm{ 1}}{{\rm{0}}^{{\rm{21}}}}$
D. $1.47{\rm{ }} \times {\rm{ }}{10^{22}}$

Answer 1

First calculate the volume of sheet from given dimensions. Then multiply the volume of the gold sheet with its density to calculate its mass. Then divide mass with atomic weight to calculate the number of moles of gold. Finally multiply the number of moles of gold with Avogadro’s number to calculate the number of gold atoms present in the gold sheet.

Complete answer:
You are given the length and breadth and thickness of a gold foil. You are also given the density of gold and its atomic weight. You are asked to calculate the number of atoms of gold present in the sheet.
The breadth and length of the gold sheet are ${\rm{25}}{\rm{.0 \,mm}}$ and ${\rm{40}}{\rm{.0 \,mm}}$ respectively. The thickness of the gold sheet is ${\rm{0}}{\rm{.25 \,mm}}$. The density of gold is ${\rm{19}}{\rm{.32 \,g/c}}{{\rm{m}}^{\rm{3}}}$ .
From the dimensions of the gold sheet calculate the volume of the gold sheet.

$$\Rightarrow {\rm{V = 25}}{\rm{.0 \,mm \times 40}}{\rm{.0 \,mm}} \times {\rm{0}}{\rm{.25 \,mm}} \\\ \Rightarrow {\rm{V = 250 \,m}}{{\rm{m}}^3}$$

Here B, L and T are volume, breadth, length and thickness respectively.
Convert the unit of volume from cubic millimeter to cubic centimeter

$$\Rightarrow {\rm{V = 0}}{\rm{.250 \,c}}{{\rm{m}}^3}{\rm{ }}$$

Multiply the volume of sheet with its density to calculate its mass

$${\rm{w = }}\rho \times {\rm{V}} \\\ \Rightarrow {\rm{w = 19}}{\rm{.32 \,g/c}}{{\rm{m}}^{\rm{3}}} \times {\rm{0}}{\rm{.250 \,c}}{{\rm{m}}^3}{\rm{ }} \\\ \Rightarrow {\rm{w = 4}}{\rm{.83 \,g}}$$

Here, w is the mass of the gold foil and $\rho$ is its density.
Divide mass of gold sheet with its atomic weight to calculate the number of moles of gold

$${\rm{n = }}\dfrac{{\rm{w}}}{{\rm{M}}} \\\ \Rightarrow {\rm{n = }}\dfrac{{{\rm{4}}{\rm{.83\, g}}}}{{{\rm{197}}{\rm{.0 \,g/mol}}}} \\\ \Rightarrow {\rm{n = }}\dfrac{{483}}{{19700}}{\rm{ mol}}$$

Multiply the number of moles of gold with Avogadro’s number to calculate the number of gold atoms

$${\rm{N = n}} \times {{\rm{N}}_{\rm{A}}} \\\ \Rightarrow {\rm{N = }}\dfrac{{483}}{{19700}}{\rm{ mol}} \times 6.023 \times {10^{23}}{\rm{ atoms/mol}} \\\ \Rightarrow {\rm{N = 1}}{\rm{.47}} \times {\rm{1}}{{\rm{0}}^{22}} \\\$$

Hence, there are $1.47{\rm{ }} \times {\rm{ }}{10^{22}}$ atoms of gold in the sheet.

**Hence, the correct option is the option (D) $1.47{\rm{ }} \times {\rm{ }}{10^{22}}$

Note:**
Avogadro’s number represents the number of gold atoms present in one mole of gold. The mass of one mole of gold is equal to its atomic weight. Thus, ${\rm{197}}{\rm{.0 grams}}$ of gold contains ${\rm{6}}{\rm{.023 \times 1}}{{\rm{0}}^{{\rm {23}}}}$ gold atoms.