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 × 10²³ mol^–1

To find the density of the metal, we use the formula for the density ($d$) of a crystalline solid:
$d=\frac{Z\times M}{a^3\times N_A}$

Let us identify the given values:
1. Type of lattice: Cubic close packed (ccp), which is equivalent to a face-centered cubic (fcc) lattice. Therefore, the number of atoms per unit cell ($Z$) is 4.
2. Molar mass of the metal ($M$): The atomic mass is given as 105.6 amu, which corresponds to a molar mass of 105.6 g mol^-1.
3. Edge length ($a$): 400 pm = 400 × 10^-12 m = 400 × 10^-10 cm = 4 × 10^-8 cm.
4. Avogadro's constant ($N_A$): 6 × 10²³ mol^-1.

Now, let us calculate the volume of the unit cell ($a^3$):
$a^3={(4\times {10}^{-8}\text{cm})}^3=64\times {10}^{-24}{\text{cm}}^3$

Substitute all the values into the density formula:
$d=\frac{4\times 105.6}{64\times {10}^{-24}\times 6\times {10}^{23}}$

Simplify the denominator:
$64\times {10}^{-24}\times 6\times {10}^{23}=384\times {10}^{-1}=38.4$

Now, calculate the final value of the density:
$d=\frac{422.4}{38.4}=11{\text{g cm}}^{-3}$

Thus, the density of the metal is 11.00 g cm^-3.