The face centered cubic cell of platinum ha an edge length of 0.392nm. Calculate the density of platinum `(g//cm^(3))` : `("Atomic mass": Pt=195)`

Options

A. 20.9
B. 20.4
C. 19.6
D. 21.5

Answer

To calculate the density of platinum in a face-centered cubic (FCC) unit cell, we can follow these steps:

### Step 1: Determine the number of atoms per unit cell (Z)
In a face-centered cubic (FCC) structure, there are 4 atoms per unit cell. This is calculated as follows:
- Each corner atom contributes \( \frac{1}{8} \) of an atom to the unit cell and there are 8 corners.
- Each face atom contributes \( \frac{1}{2} \) of an atom to the unit cell and there are 6 faces.

Thus, the total number of atoms \( Z \) in the FCC unit cell is:
\[ Z = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4 \]

### Step 2: Calculate the volume of the unit cell
The volume \( V \) of the unit cell can be calculated using the edge length \( a \):
\[ V = a^3 \]
Given that the edge length \( a = 0.392 \, \text{nm} = 0.392 \times 10^{-7} \, \text{cm} \):
\[ V = (0.392 \times 10^{-7})^3 \, \text{cm}^3 \]

### Step 3: Calculate the mass of the unit cell
The mass \( m \) of the unit cell can be calculated using the formula:
\[ m = Z \times \text{(Atomic mass)} / N_A \]
Where:
- \( Z = 4 \) (from Step 1)
- Atomic mass of platinum \( = 195 \, \text{g/mol} \)
- \( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \) (Avogadro's number)

Substituting the values:
\[ m = 4 \times \frac{195}{6.022 \times 10^{23}} \, \text{g} \]

### Step 4: Calculate the density
Density \( \rho \) is given by:
\[ \rho = \frac{m}{V} \]
Substituting the values of mass and volume from Steps 2 and 3:
\[ \rho = \frac{4 \times \frac{195}{6.022 \times 10^{23}}}{(0.392 \times 10^{-7})^3} \]

### Step 5: Perform the calculations
1. Calculate the volume:
\[ V = (0.392 \times 10^{-7})^3 = 6.036 \times 10^{-23} \, \text{cm}^3 \]

2. Calculate the mass:
\[ m = 4 \times \frac{195}{6.022 \times 10^{23}} = 1.295 \times 10^{-21} \, \text{g} \]

3. Calculate the density:
\[ \rho = \frac{1.295 \times 10^{-21}}{6.036 \times 10^{-23}} \approx 21.49 \, \text{g/cm}^3 \]

Thus, the density of platinum is approximately \( 21.5 \, \text{g/cm}^3 \).