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. `"10 g cm"^(-3)`
B. `" 14.2 g cm"^(-3)`
C. `"6.15 g cm"^(-3)`
D. `"9.32 g cm"^(-3)`

Explanation

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

\[ \rho = \frac{Z \cdot M}{N_A \cdot a^3} \]

Where:
- \(\rho\) = density of the metal (g/cm³)
- \(Z\) = number of atoms per unit cell (for BCC, \(Z = 2\))
- \(M\) = molar mass of the metal (g/mol)
- \(N_A\) = Avogadro's number (\(6.022 \times 10^{23} \text{ mol}^{-1}\))
- \(a\) = edge length of the unit cell (cm)

### Step 1: Convert the edge length from picometers to centimeters.
Given:
- \(a = 300 \text{ pm} = 300 \times 10^{-12} \text{ m} = 300 \times 10^{-10} \text{ cm} = 3.00 \times 10^{-8} \text{ cm}\)

### Step 2: Substitute the values into the density formula.
We know:
- \(Z = 2\) (for BCC)
- \(M = 50 \text{ g/mol}\)
- \(N_A = 6.022 \times 10^{23} \text{ mol}^{-1}\)
- \(a = 3.00 \times 10^{-8} \text{ cm}\)

Now substituting these values into the density formula:

\[ \rho = \frac{2 \cdot 50}{6.022 \times 10^{23} \cdot (3.00 \times 10^{-8})^3} \]

### Step 3: Calculate \(a^3\).
Calculating \(a^3\):

\[ (3.00 \times 10^{-8})^3 = 27.00 \times 10^{-24} \text{ cm}^3 = 2.7 \times 10^{-23} \text{ cm}^3 \]

### Step 4: Substitute \(a^3\) back into the density formula.
Now substituting \(a^3\) back into the equation:

\[ \rho = \frac{100}{6.022 \times 10^{23} \cdot 2.7 \times 10^{-23}} \]

### Step 5: Calculate the denominator.
Calculating the denominator:

\[ 6.022 \times 10^{23} \cdot 2.7 \times 10^{-23} = 16.2274 \]

### Step 6: Calculate the density.
Now substituting back into the density formula:

\[ \rho = \frac{100}{16.2274} \approx 6.15 \text{ g/cm}^3 \]

### Final Answer:
Thus, the density of the metal is approximately **6.15 g/cm³**.

---