A metal of density `7.5 xx 10^(3) kg m^(-3)` has an fcc crystal structure with lattice parameter `a = 400 pm`. Calculater the number of unit cells present in `0.015 kg` of metal.

Options

A. `6.250 xx 10^(22)`
B. `3.125 xx 10^(23)`
C. `3.125 xx 10^(22)`
D. `1.563 xx 10^(22)`

Explanation

To calculate the number of unit cells present in 0.015 kg of a metal with a face-centered cubic (FCC) structure and given density, we can follow these steps:

### Step 1: Understand the given data
- Density (d) = \( 7.5 \times 10^3 \, \text{kg/m}^3 \)
- Lattice parameter (a) = 400 pm = \( 400 \times 10^{-12} \, \text{m} \)
- Mass of the metal (m) = 0.015 kg
- For FCC structure, the number of atoms per unit cell (Z) = 4.

### Step 2: Calculate the volume of one unit cell
The volume (V) of a cubic unit cell can be calculated using the formula:
\[ V = a^3 \]
Substituting the value of \( a \):
\[ V = (400 \times 10^{-12})^3 = 64 \times 10^{-36} \, \text{m}^3 \]

### Step 3: Calculate the mass of one unit cell
The mass (m_cell) of one unit cell can be calculated using the formula:
\[ \text{mass of one unit cell} = \text{density} \times \text{volume} \]
Substituting the values:
\[ m_{\text{cell}} = d \times V = (7.5 \times 10^3) \times (64 \times 10^{-36}) = 4.8 \times 10^{-32} \, \text{kg} \]

### Step 4: Calculate the number of unit cells
The total number of unit cells (n) can be calculated by dividing the total mass of the metal by the mass of one unit cell:
\[ n = \frac{\text{total mass}}{\text{mass of one unit cell}} = \frac{0.015}{4.8 \times 10^{-32}} \]
Calculating this gives:
\[ n = \frac{0.015}{4.8 \times 10^{-32}} \approx 3.125 \times 10^{22} \]

### Final Answer
The number of unit cells present in 0.015 kg of the metal is approximately \( 3.125 \times 10^{22} \).

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