1. Identify the crystal type

For a cubic close packed structure, the lattice is face-centered cubic (fcc).

So, number of atoms per unit cell: $Z = 4$

2. Use density formula

Density of a unit cell is given by: $\rho = \frac{Z \cdot M}{N_A \cdot a^3}$ where:

• $Z = 4$
• $M = 105.6\,\text{g mol}^{-1}$
• $N_A = 6 \times 10^{23}\,\text{mol}^{-1}$
• $a = 400\,\text{pm}$

3. Convert edge length into cm

Since: $1\,\text{pm} = 10^{-10}\,\text{cm}$ therefore, $a = 400\,\text{pm} = 400 \times 10^{-10}\,\text{cm} = 4 \times 10^{-8}\,\text{cm}$

4. Calculate unit cell volume

$a^3 = (4 \times 10^{-8})^3 = 64 \times 10^{-24} = 6.4 \times 10^{-23}\,\text{cm}^3$

5. Calculate mass of one unit cell

$\text{Mass of unit cell} = \frac{Z \cdot M}{N_A} = \frac{4 \times 105.6}{6 \times 10^{23}}$

$= \frac{422.4}{6 \times 10^{23}} = 70.4 \times 10^{-23} = 7.04 \times 10^{-22}\,\text{g}$

6. Compute density

$\rho = \frac{7.04 \times 10^{-22}}{6.4 \times 10^{-23}}$

$\rho = 11\,\text{g cm}^{-3}$

7. Final answer

$\boxed{11}$

8. Comparison with stored answer

Stored correct answer range is $10.85$ to $11.1$.

Our calculated answer $11$ lies within this range, so it agrees.