.16: Calculate the density of iron, given that it has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomic mass of 55.9 g/mole

8 g/cm³

Explanation:

So, we are given the following information or parameters which are going to help in solving thus particular question/problem;

We are given that the atomic radius for the BCC crystal structure = 0.124 nm, and it has an atomic mass which is equal to = 55.9 g/mole.

We are to determine or Calculate the density of iron from the given data or information above. So, we will be making use of the formula below:

The density of iron =[ number of atoms or unit cells × atomic mass or atomic weight] ÷ [ 4 × atomic radius/√3]³ × Avogadro's number.

The number of atoms or unit cells = 2, atomic mass or atomic weight =  55.9 g/mole.

Thus, the density of iron =[ 2 × 55.9]/[ 4× 0.214 × 10^-7/√3]³ × 6.022 × 10^23] = 8 g/cm³.

the answer is c.the valence of the atoms combine .