A metal crystallises as body centred cubic lattice with the edge length of unit cell equal to 0.304 nm. If the molar mass of the metal is 50.3 g` mol^(−1)` , its density (in g `cm^(−3)` ) is :

Options

A. 5.945
B. 2.9725
C. 8.915
D. 4.458

Answer

To calculate the density of the metal that crystallizes in a body-centered cubic (BCC) lattice, we can use the formula for density:

\[ \text{Density} (d) = \frac{Z \cdot M}{a^3 \cdot N_A} \]

Where:
- \( Z \) = number of atoms per unit cell (for BCC, \( Z = 2 \))
- \( M \) = molar mass of the metal (given as 50.3 g/mol)
- \( a \) = edge length of the unit cell (given as 0.304 nm)
- \( N_A \) = Avogadro's number (\( 6.02 \times 10^{23} \) mol\(^{-1}\))

### Step-by-Step Solution:

**Step 1: Convert the edge length from nanometers to centimeters.**
- Given edge length \( a = 0.304 \) nm.
- To convert to centimeters:
\[ a = 0.304 \, \text{nm} = 0.304 \times 10^{-7} \, \text{cm} \]

**Step 2: Calculate \( a^3 \).**
- Now calculate \( a^3 \):
\[ a^3 = (0.304 \times 10^{-7})^3 = 0.304^3 \times 10^{-21} \, \text{cm}^3 \]
Calculating \( 0.304^3 \):
\[ 0.304^3 \approx 0.0283 \]
Thus,
\[ a^3 \approx 0.0283 \times 10^{-21} \, \text{cm}^3 \]

**Step 3: Substitute values into the density formula.**
- Now substitute \( Z = 2 \), \( M = 50.3 \, \text{g/mol} \), \( a^3 \approx 0.0283 \times 10^{-21} \, \text{cm}^3 \), and \( N_A = 6.02 \times 10^{23} \):
\[ d = \frac{2 \cdot 50.3}{0.0283 \times 10^{-21} \cdot 6.02 \times 10^{23}} \]

**Step 4: Calculate the denominator.**
- Calculate \( a^3 \cdot N_A \):
\[ 0.0283 \times 10^{-21} \cdot 6.02 \times 10^{23} \approx 0.1704 \, \text{cm}^3 \]

**Step 5: Calculate the density.**
- Now calculate the density:
\[ d = \frac{2 \cdot 50.3}{0.1704} \approx \frac{100.6}{0.1704} \approx 5.91 \, \text{g/cm}^3 \]

### Final Result:
The density of the metal is approximately \( 5.91 \, \text{g/cm}^3 \).