Density of a unit cell is respresented as
`rho = (" Effective no. of atoms (s)" xx "Mass of a unit cell ")/("Volume of a unit cell ")=(Z.M)/(N_(A).a^(3))`
where , Z =mass of effective no . of atoms(s) or ion (s).
M= At . mass// formula
`N_(A) `= Avogadro' s no . `rArr 6.0323 xx 10^(23)`
a= edge length of unit cell
An element crystallizes in a structure having fcc unit cell of an edge 200 pm . Calculate the density , if 100 g of this element contains `12xx10^(23)` atoms :

Options

A. `41.66 g //cm^(3)`
B. `4.166 g// cm^(3)`
C. `10.25 g // cm^(3)`
D. `1.025 g // cm^(3)`

Answer

To solve the problem of calculating the density of an element that crystallizes in a face-centered cubic (FCC) unit cell, we will follow these steps:

### Step 1: Identify the Given Data
- Edge length of the unit cell (a) = 200 pm = \(200 \times 10^{-12}\) m
- Number of atoms in 100 g = \(12 \times 10^{23}\) atoms
- Avogadro's number (\(N_A\)) = \(6.022 \times 10^{23}\) atoms/mol

### Step 2: Calculate the Atomic Mass (M)
To find the atomic mass (M) of the element, we first determine the number of atoms per gram:
\[ \text{Mass of one atom} = \frac{100 \text{ g}}{12 \times 10^{23} \text{ atoms}} = \frac{100}{12 \times 10^{23}} \text{ g/atom} \]

Now, to find the mass of one mole (which is the atomic mass):
\[ M = \text{Mass of one atom} \times N_A = \left(\frac{100}{12 \times 10^{23}}\right) \times (6.022 \times 10^{23}) \text{ g} \]
\[ M = \frac{100 \times 6.022}{12} \text{ g} = 50.18 \text{ g/mol} \]

### Step 3: Determine the Effective Number of Atoms (Z) in FCC
In an FCC unit cell, the effective number of atoms (Z) is 4. This is because:
- There are 8 corner atoms, each contributing \( \frac{1}{8} \) to the unit cell.
- There are 6 face-centered atoms, each contributing \( \frac{1}{2} \) to the unit cell.

Calculating:
\[ Z = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4 \]

### Step 4: Calculate the Volume of the Unit Cell
The volume of the unit cell (V) is given by:
\[ V = a^3 \]
Substituting the edge length:
\[ V = (200 \times 10^{-12} \text{ m})^3 = (2 \times 10^{-10} \text{ m})^3 = 8 \times 10^{-30} \text{ m}^3 \]

To convert to cm³:
\[ V = 8 \times 10^{-30} \text{ m}^3 \times (100^3) \text{ cm}^3/\text{m}^3 = 8 \times 10^{-24} \text{ cm}^3 \]

### Step 5: Calculate the Density (ρ)
Using the density formula:
\[ \rho = \frac{Z \cdot M}{N_A \cdot V} \]
Substituting the values:
\[ \rho = \frac{4 \cdot 50.18 \text{ g/mol}}{6.022 \times 10^{23} \text{ atoms/mol} \cdot 8 \times 10^{-24} \text{ cm}^3} \]
Calculating:
\[ \rho = \frac{200.72}{6.022 \times 10^{23} \cdot 8 \times 10^{-24}} = \frac{200.72}{4.8176} \approx 41.66 \text{ g/cm}^3 \]

### Final Answer
The density of the element is approximately \(41.66 \text{ g/cm}^3\).

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