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

Options

A. `1times10^(25)`
B. `1times10^(24)`
C. `1times10^(22)`
D. `1times10^(20)`

Explanation

To solve the problem, we need to find the number of unit cells in 200 grams of a metallic element with a cubic lattice, given the edge length and density. Here’s a step-by-step solution:

### Step 1: Convert the edge length to centimeters
The edge length of the unit cell is given as 2 Å (angstroms). We need to convert this to centimeters.

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

### Step 2: Calculate the volume of the unit cell
The volume \( V \) of a cubic unit cell can be calculated using the formula:

\[ V = A^3 \]

Substituting the value of \( A \):

\[ V = (2 \times 10^{-8} \, \text{cm})^3 = 8 \times 10^{-24} \, \text{cm}^3 \]

### Step 3: Calculate the mass of the unit cell
The mass \( m \) of the unit cell can be calculated using the formula:

\[ m = \text{Volume} \times \text{Density} \]

Given the density is 2.5 g/cm³:

\[ m = 8 \times 10^{-24} \, \text{cm}^3 \times 2.5 \, \text{g/cm}^3 = 2 \times 10^{-23} \, \text{g} \]

### Step 4: Calculate the number of unit cells in 200 grams of the metal
To find the number of unit cells \( N \) in 200 grams, we use the formula:

\[ N = \frac{\text{Total mass}}{\text{Mass of one unit cell}} \]

Substituting the values:

\[ N = \frac{200 \, \text{g}}{2 \times 10^{-23} \, \text{g}} = 1 \times 10^{25} \]

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

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