Calculate Density of Metal in CCP Lattice

To calculate the density of the metal, we use the standard unit cell density formula:

$\rho = \frac{Z \times M}{a^3 \times N_A}$

where:

• $Z$ is the number of atoms per unit cell. For a cubic close packed (ccp) or face-centered cubic (fcc) lattice, $Z = 4$.
• $M$ is the atomic mass of the metal $= 105.6 \text{ g mol}^{-1}$.
• $a$ is the edge length of the unit cell $= 400 \text{ pm} = 400 \times 10^{-10} \text{ cm} = 4 \times 10^{-8} \text{ cm}$.
• $N_A$ is Avogadro's constant $= 6 \times 10^{23} \text{ mol}^{-1}$.

Step-by-Step Calculation:

1. Volume of the unit cell ($V = a^3$): $V = (4 \times 10^{-8} \text{ cm})^3 = 64 \times 10^{-24} \text{ cm}^3$

2. Substitute the given values into the density equation: $\rho = \frac{4 \times 105.6 \text{ g mol}^{-1}}{(64 \times 10^{-24} \text{ cm}^3) \times (6 \times 10^{23} \text{ mol}^{-1})}$

3. Simplify the denominator: $\text{Denominator} = 64 \times 6 \times 10^{-1} = 384 \times 10^{-1} = 38.4 \text{ cm}^3 \text{ mol}^{-1}$

4. Calculate the density ($\rho$): $\rho = \frac{422.4}{38.4} = 11 \text{ g cm}^{-3}$

The density of the metal is 11 (or 11.00).