An element crystallizes in f.c.c. lattice with edge length of 400 pm. The density of the element is 7 g `cm^(-3)`. How many atoms are present in 280 g of the element ?

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Understand the given data
- The element crystallizes in a face-centered cubic (FCC) lattice.
- Edge length (a) = 400 pm = \(400 \times 10^{-10}\) cm
- Density (D) = 7 g/cm³
- Mass of the element = 280 g

### Step 2: Calculate the volume of the unit cell
The volume (V) of the unit cell can be calculated using the formula:
\[ V = a^3 \]
Substituting the value of edge length:
\[ V = (400 \times 10^{-10} \text{ cm})^3 = 64 \times 10^{-30} \text{ cm}^3 = 6.4 \times 10^{-29} \text{ cm}^3 \]

### Step 3: Calculate the molar mass (M) using the density formula
The density formula relates density, mass, and volume:
\[ D = \frac{z \cdot M}{V_m} \]
Where:
- \(D\) = density
- \(z\) = number of atoms per unit cell (for FCC, \(z = 4\))
- \(M\) = molar mass
- \(V_m\) = volume of the unit cell in cm³

Rearranging the formula to find molar mass:
\[ M = \frac{D \cdot V_m}{z} \]
Substituting the known values:
\[ M = \frac{7 \text{ g/cm}^3 \cdot 6.4 \times 10^{-29} \text{ cm}^3}{4} = \frac{44.8 \times 10^{-29} \text{ g}}{4} = 11.2 \times 10^{-29} \text{ g} = 67.4 \text{ g/mol} \]

### Step 4: Calculate the number of moles in 280 g of the element
Using the formula for moles:
\[ \text{Moles} = \frac{\text{mass}}{\text{molar mass}} \]
Substituting the values:
\[ \text{Moles} = \frac{280 \text{ g}}{67.4 \text{ g/mol}} \approx 4.15 \text{ mol} \]

### Step 5: Calculate the number of atoms
To find the number of atoms, we use Avogadro's number (\(N_A = 6.022 \times 10^{23} \text{ atoms/mol}\)):
\[ \text{Number of atoms} = \text{Moles} \times N_A \]
Substituting the values:
\[ \text{Number of atoms} = 4.15 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} \approx 2.50 \times 10^{24} \text{ atoms} \]

### Final Answer
The number of atoms present in 280 g of the element is approximately \(2.50 \times 10^{24}\) atoms.

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