The density (in g cm$$^{-3}$$) of the metal which forms a cubic close packed (ccp) lattice with an axial distance (edge length) equal to 400 pm is ______.

Use: Atomic mass of metal = 105.6 amu and Avogadro's constant = $$6 \times 10^{23}$$ mol$$^{-1}$$

For any crystalline solid, density is obtained from a single unit cell using the relation

$$\rho = \frac{Z \, M}{N_A \, a^{3}}$$
where
  $$Z$$ = number of atoms per unit cell,
  $$M$$ = molar mass (g mol$$^{-1}$$),
  $$N_A$$ = Avogadro’s constant (mol$$^{-1}$$),
  $$a$$ = edge length of the cubic unit cell (cm).

Step 1: Identify the parameters for the given lattice.
A cubic close-packed (ccp) lattice is face-centred cubic (fcc), so $$Z = 4$$.

Step 2: Convert the edge length to centimetres.
Given $$a = 400 \text{ pm}$$.
$$1 \text{ pm} = 10^{-12} \text{ m} = 10^{-10} \text{ cm}$$.
Hence $$a = 400 \times 10^{-10} \text{ cm} = 4.0 \times 10^{-8} \text{ cm}$$.

Step 3: Cube the edge length.
$$a^{3} = (4.0 \times 10^{-8}\,\text{cm})^{3} = 4^{3} \times 10^{-24}\,\text{cm}^{3} = 64 \times 10^{-24}\,\text{cm}^{3} = 6.4 \times 10^{-23}\,\text{cm}^{3}$$.

Step 4: Substitute all values into the density formula.
Numerator: $$Z \, M = 4 \times 105.6 = 422.4 \text{ g mol}^{-1}$$.
Denominator: $$N_A \, a^{3} = (6 \times 10^{23}) \times (6.4 \times 10^{-23}) = 38.4$$.

Therefore,
$$\rho = \frac{422.4}{38.4} = 11.0 \text{ g cm}^{-3}$$.

The calculated density (to three significant figures) is $$11.0 \text{ g cm}^{-3}$$, which lies in the required range 10.85-11.1.