The density of mercury is `13.6 g mL^(-1)`. Calculate the approximate diameter of an atom of mercury assuming that each atom is occupying a cube of edge length equal to the diameter of the mercury atom.

Options

A. 3.01 Å
B. 2.54 Å
C. 0.29 Å
D. 2.91 Å

Explanation

To calculate the approximate diameter of an atom of mercury given its density, we can follow these steps:

### Step 1: Understand the given data
- Density of mercury (ρ) = 13.6 g/mL
- Molecular mass of mercury (M) = 200 g/mol
- Avogadro's number (N_A) = 6.022 × 10²³ atoms/mol

### Step 2: Convert density to g/cm³
Since 1 mL = 1 cm³, the density can be directly used as:
- ρ = 13.6 g/cm³

### Step 3: Calculate the number of atoms in 1 g of mercury
To find the number of atoms in 1 g of mercury, we use the formula:
\[ \text{Number of atoms} = \frac{N_A}{M} \]
For 1 g of mercury:
\[ \text{Number of atoms in 1 g} = \frac{6.022 \times 10^{23} \text{ atoms/mol}}{200 \text{ g/mol}} = 3.011 \times 10^{21} \text{ atoms} \]

### Step 4: Calculate the volume of 1 g of mercury
The volume (V) of 1 g of mercury can be calculated using the density formula:
\[ V = \frac{\text{mass}}{\text{density}} \]
For 1 g of mercury:
\[ V = \frac{1 \text{ g}}{13.6 \text{ g/cm}^3} = 0.07353 \text{ cm}^3 \]

### Step 5: Calculate the volume occupied by one atom of mercury
The volume occupied by one atom can be calculated by dividing the total volume by the number of atoms:
\[ V_{\text{atom}} = \frac{V}{\text{Number of atoms}} = \frac{0.07353 \text{ cm}^3}{3.011 \times 10^{21} \text{ atoms}} \]
Calculating this gives:
\[ V_{\text{atom}} \approx 2.44 \times 10^{-23} \text{ cm}^3 \]

### Step 6: Relate the volume of one atom to its edge length
Assuming each atom occupies a cube of edge length \( a \):
\[ V_{\text{atom}} = a^3 \]
Thus,
\[ a = \sqrt[3]{V_{\text{atom}}} = \sqrt[3]{2.44 \times 10^{-23} \text{ cm}^3} \]

### Step 7: Calculate the cube root
Calculating the cube root gives:
\[ a \approx 2.905 \times 10^{-8} \text{ cm} \]

### Step 8: Convert to angstroms
To convert centimeters to angstroms (1 cm = 10^{8} Å):
\[ \text{Diameter} = a \approx 2.905 \times 10^{-8} \text{ cm} \times 10^{8} \text{ Å/cm} = 2.905 \text{ Å} \]

### Final Result
The approximate diameter of a mercury atom is:
\[ \text{Diameter} \approx 2.91 \text{ Å} \]