A `25.0 mm xx 40.0mm` piece of gold foil is `0.25mm` thick. The density of gold is `19.32g/cm^3`. How many gold atoms are in the sheet ? (Atomic weight : `Au=197.0`)

Options

A. `7.7xx10^(23)`
B. `1.5 xx10^(23)`
C. `4.3 xx10^(21)`
D. `1.47xx10^(22)`

Answer

To solve the problem of finding how many gold atoms are in a piece of gold foil with given dimensions and density, we can follow these steps:

### Step 1: Calculate the Volume of the Gold Foil
The volume \( V \) of the gold foil can be calculated using the formula:
\[ V = \text{length} \times \text{breadth} \times \text{thickness} \]
Given:
- Length = \( 25.0 \, \text{mm} \)
- Breadth = \( 40.0 \, \text{mm} \)
- Thickness = \( 0.25 \, \text{mm} \)

Calculating the volume:
\[ V = 25.0 \, \text{mm} \times 40.0 \, \text{mm} \times 0.25 \, \text{mm} = 250 \, \text{mm}^3 \]

### Step 2: Convert Volume to cm³
Since the density is given in \( \text{g/cm}^3 \), we need to convert the volume from mm³ to cm³. The conversion factor is:
\[ 1 \, \text{cm}^3 = 1000 \, \text{mm}^3 \]
Thus,
\[ V = 250 \, \text{mm}^3 \times \frac{1 \, \text{cm}^3}{1000 \, \text{mm}^3} = 0.250 \, \text{cm}^3 \]

### Step 3: Calculate the Mass of the Gold Foil
Using the density of gold, we can calculate the mass \( m \) using the formula:
\[ m = \text{density} \times \text{volume} \]
Given:
- Density of gold = \( 19.32 \, \text{g/cm}^3 \)

Calculating the mass:
\[ m = 19.32 \, \text{g/cm}^3 \times 0.250 \, \text{cm}^3 = 4.83 \, \text{g} \]

### Step 4: Calculate the Number of Moles of Gold
To find the number of moles \( n \), we use the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} \]
Given:
- Molar mass of gold (Au) = \( 197.0 \, \text{g/mol} \)

Calculating the number of moles:
\[ n = \frac{4.83 \, \text{g}}{197.0 \, \text{g/mol}} \approx 0.0245 \, \text{moles} \]

### Step 5: Calculate the Number of Atoms in the Gold Foil
Using Avogadro's number \( N_A = 6.022 \times 10^{23} \, \text{atoms/mol} \), we can find the number of atoms \( N \):
\[ N = n \times N_A \]
Calculating the number of atoms:
\[ N = 0.0245 \, \text{moles} \times 6.022 \times 10^{23} \, \text{atoms/mol} \approx 1.47 \times 10^{22} \, \text{atoms} \]

### Final Answer
The number of gold atoms in the sheet is approximately \( 1.47 \times 10^{22} \, \text{atoms} \).

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