An element occurs in bcc structure with cell edge 288 pm. Its density is 7.2 g `cm^(-3)`. Calculate the atomic mass of the element.

Explanation

To calculate the atomic mass of the element that occurs in a body-centered cubic (BCC) structure, we can follow these steps:

### Step 1: Convert the cell edge length from picometers to centimeters
The given cell edge length is 288 pm (picometers). We need to convert this to centimeters.

\[ \text{Cell edge length (a)} = 288 \, \text{pm} = 288 \times 10^{-12} \, \text{m} = 288 \times 10^{-10} \, \text{cm} \]

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

\[ V = a^3 \]

Substituting the value of \( a \):

\[ V = (288 \times 10^{-10} \, \text{cm})^3 \]

Calculating this gives:

\[ V = 2.39 \times 10^{-29} \, \text{cm}^3 \]

### Step 3: Use the density formula to find the mass of the unit cell
The density \( d \) of the element is given as 7.2 g/cm³. The mass \( m \) of the unit cell can be calculated using the formula:

\[ d = \frac{m}{V} \]

Rearranging this gives:

\[ m = d \times V \]

Substituting the values:

\[ m = 7.2 \, \text{g/cm}^3 \times 2.39 \times 10^{-29} \, \text{cm}^3 \]

Calculating this gives:

\[ m = 1.72 \times 10^{-28} \, \text{g} \]

### Step 4: Calculate the molar mass using Avogadro's number
In a BCC structure, there are 2 atoms per unit cell (z = 2). To find the molar mass \( M \) of the element, we can use the relationship:

\[ M = \frac{m \times N_A}{z} \]

Where \( N_A \) (Avogadro's number) is approximately \( 6.022 \times 10^{23} \, \text{mol}^{-1} \).

Substituting the values:

\[ M = \frac{1.72 \times 10^{-28} \, \text{g} \times 6.022 \times 10^{23} \, \text{mol}^{-1}}{2} \]

Calculating this gives:

\[ M = 51.76 \, \text{g/mol} \]

### Final Answer
The atomic mass of the element is approximately **51.76 g/mol**.

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