The elements crystallizes in a body centered cubic lattice and the edge of the unit cell is 0.351nm. The density is `0.533g//cm^(3)`. What is the atomic mas?

Options

A. 12
B. 6.94
C. 9.01
D. 10.8

Answer

To find the atomic mass of the element that crystallizes in a body-centered cubic (BCC) lattice, we can use the formula for density in relation to the unit cell parameters. Here’s a step-by-step solution:

### Step 1: Understand the relationship between density, mass, and volume
The formula for density (ρ) is given by:

\[ \rho = \frac{Z \cdot M}{V} \]

where:
- \( \rho \) = density (g/cm³)
- \( Z \) = number of atoms per unit cell
- \( M \) = molar mass (g/mol)
- \( V \) = volume of the unit cell (cm³)

### Step 2: Identify parameters for BCC
For a body-centered cubic (BCC) lattice:
- \( Z = 2 \) (there are 2 atoms per unit cell)

### Step 3: Calculate the volume of the unit cell
The volume \( V \) of the unit cell can be calculated using the edge length \( a \):

\[ V = a^3 \]

Given that the edge length \( a = 0.351 \, \text{nm} = 0.351 \times 10^{-7} \, \text{cm} \):

\[ V = (0.351 \times 10^{-7} \, \text{cm})^3 = 4.33 \times 10^{-22} \, \text{cm}^3 \]

### Step 4: Substitute values into the density formula
Now, substituting the known values into the density formula:

\[ 0.533 = \frac{2 \cdot M}{4.33 \times 10^{-22}} \]

### Step 5: Solve for M (molar mass)
Rearranging the equation to solve for \( M \):

\[ M = \frac{0.533 \cdot 4.33 \times 10^{-22}}{2} \]

Calculating \( M \):

\[ M = \frac{0.533 \cdot 4.33 \times 10^{-22}}{2} = 1.15 \times 10^{-22} \, \text{g/mol} \]

### Step 6: Convert to a more usable unit
To find the molar mass in grams per mole, we need to multiply by Avogadro's number \( (6.022 \times 10^{23} \, \text{mol}^{-1}) \):

\[ M = 1.15 \times 10^{-22} \cdot 6.022 \times 10^{23} = 6.93 \, \text{g/mol} \]

### Final Answer
The atomic mass of the element is approximately **6.93 g/mol**.

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