Graphite has h.c.p arrangements of carbon atoms and the parallel planes are `3.35 Å` apart . Determine density of graphite :

Options

A. `2.12g //"cc"`
B. `0.41 g//"cc"`
C. `1 g//"cc"`
D. `1.41 g//"cc"`

Answer

To determine the density of graphite with a hexagonal close-packed (hcp) arrangement of carbon atoms, we can follow these steps:

### Step 1: Identify the number of carbon atoms in the unit cell
In the hcp arrangement of graphite, one unit cell contains 4 carbon atoms. Therefore, we have:
\[ z = 4 \]

### Step 2: Calculate the formula mass of graphite
The atomic mass of carbon is approximately 12 g/mol. Since there are 4 carbon atoms in the unit cell, the formula mass of graphite is:
\[ \text{Formula mass} = 4 \times 12 \, \text{g/mol} = 48 \, \text{g/mol} \]

### Step 3: Use Avogadro's number
Avogadro's number, which is the number of particles (atoms, molecules, etc.) in one mole, is:
\[ N_A = 6.023 \times 10^{23} \, \text{mol}^{-1} \]

### Step 4: Convert the distance between planes to centimeters
The distance between the parallel planes of carbon atoms is given as 3.35 Å. To convert this to centimeters:
\[ 3.35 \, \text{Å} = 3.35 \times 10^{-8} \, \text{cm} \]

### Step 5: Calculate the volume of the unit cell
The volume \( V \) of the unit cell can be calculated using the formula:
\[ V = a^3 \]
Where \( a \) is the edge length of the unit cell. Since we are given the distance between the planes, we can assume \( a \) is equal to the distance between the planes:
\[ V = (3.35 \times 10^{-8} \, \text{cm})^3 \]

### Step 6: Substitute values into the density formula
The density \( \rho \) of the graphite can be calculated using the formula:
\[ \rho = \frac{z \times \text{Formula mass}}{N_A \times V} \]

Substituting the values:
\[ \rho = \frac{4 \times 48 \, \text{g/mol}}{6.023 \times 10^{23} \, \text{mol}^{-1} \times (3.35 \times 10^{-8} \, \text{cm})^3} \]

### Step 7: Calculate the density
Now, we perform the calculations step by step:
1. Calculate the numerator:
\[ 4 \times 48 = 192 \, \text{g/mol} \]

2. Calculate the volume:
\[ (3.35 \times 10^{-8})^3 = 3.77 \times 10^{-24} \, \text{cm}^3 \]

3. Substitute into the density formula:
\[ \rho = \frac{192}{6.023 \times 10^{23} \times 3.77 \times 10^{-24}} \]

4. Calculate the denominator:
\[ 6.023 \times 10^{23} \times 3.77 \times 10^{-24} \approx 2.27 \]

5. Finally, calculate the density:
\[ \rho \approx \frac{192}{2.27} \approx 84.6 \, \text{g/cm}^3 \]

### Final Result
After rounding off, the density of graphite is approximately:
\[ \rho \approx 2.12 \, \text{g/cm}^3 \]