A metallic element has a cubic lattice. Each edge of the unit cell is 2 Å. The density of the metal is `2.5 g cm^(–3)` . The number of unit cells in 200 g of the metal will be:

Options

A. `1xx10^(25)`
B. `1xx10^(24)`
C. `1xx10^(22)`
D. `1xx10^(20)`

Answer

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

### Step 1: Convert the edge length of the unit cell from angstroms to centimeters.
The edge length of the unit cell is given as 2 Å (angstroms). We know that:
1 Å = \(1 \times 10^{-8}\) cm.

So,
\[ \text{Edge length (a)} = 2 \, \text{Å} = 2 \times 10^{-8} \, \text{cm} \]

### Step 2: Calculate the volume of one unit cell.
The volume \(V\) of a cubic unit cell is given by the formula:
\[ V = a^3 \]
Substituting the value of \(a\):
\[ V = (2 \times 10^{-8} \, \text{cm})^3 = 8 \times 10^{-24} \, \text{cm}^3 \]

### Step 3: Use the density to find the mass of one unit cell.
Density (\(D\)) is defined as mass (\(m\)) per unit volume (\(V\)):
\[ D = \frac{m}{V} \]
Rearranging the formula to find the mass of one unit cell:
\[ m = D \times V \]
Substituting the values:
\[ m = 2.5 \, \text{g/cm}^3 \times 8 \times 10^{-24} \, \text{cm}^3 = 20 \times 10^{-24} \, \text{g} = 2 \times 10^{-23} \, \text{g} \]

### Step 4: Calculate the number of unit cells in 200 g of the metal.
To find the number of unit cells (\(N\)), we use the formula:
\[ N = \frac{\text{Total mass}}{\text{Mass of one unit cell}} \]
Substituting the values:
\[ N = \frac{200 \, \text{g}}{2 \times 10^{-23} \, \text{g}} = 100 \times 10^{23} = 1 \times 10^{25} \]

### Final Answer:
The number of unit cells in 200 g of the metal is \(1 \times 10^{25}\).

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