A metal crystallises in a bcc lattice ,its unit cell edge length in about `300` pm and its molar mass is about `50g mol^(-1)` what would be the density of the metal (in g `cm^(-3))`?

Options

A. `3.1`
B. `6.2`
C. `9.3`
D. `12.4`

Explanation

To find the density of the metal that crystallizes in a body-centered cubic (BCC) lattice, we can follow these steps:

### Step 1: Identify the parameters
- **Unit cell edge length (a)** = 300 pm = \(300 \times 10^{-10}\) cm
- **Molar mass (M)** = 50 g/mol
- **Avogadro's number (N_A)** = \(6.022 \times 10^{23}\) mol\(^{-1}\)
- **Number of atoms per unit cell (Z)** for BCC = 2

### Step 2: Calculate the volume of the unit cell
The volume \(V\) of the unit cell can be calculated using the formula:
\[ V = a^3 \]
Substituting the value of \(a\):
\[ V = (300 \times 10^{-10} \text{ cm})^3 = 27 \times 10^{-30} \text{ cm}^3 \]

### Step 3: Calculate the mass of the unit cell
The mass \(m\) of the unit cell can be calculated using the formula:
\[ m = \frac{Z \cdot M}{N_A} \]
Substituting the values:
\[ m = \frac{2 \cdot 50 \text{ g/mol}}{6.022 \times 10^{23} \text{ mol}^{-1}} = \frac{100}{6.022 \times 10^{23}} \text{ g} \]

### Step 4: Calculate the density
Density \(\rho\) is given by the formula:
\[ \rho = \frac{m}{V} \]
Substituting the values of \(m\) and \(V\):
\[ \rho = \frac{\frac{100}{6.022 \times 10^{23}}}{27 \times 10^{-30}} \text{ g/cm}^3 \]

### Step 5: Simplify the expression
Calculating the density:
\[ \rho = \frac{100}{6.022 \times 10^{23} \cdot 27 \times 10^{-30}} \]
Calculating the denominator:
\[ 6.022 \times 27 \approx 162.594 \]
Thus,
\[ \rho \approx \frac{100}{162.594} \approx 0.615 \text{ g/cm}^3 \]

### Final Calculation
To express it in a more standard form:
\[ \rho \approx 6.15 \text{ g/cm}^3 \]

### Conclusion
The density of the metal is approximately **6.15 g/cm³**.

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