Fe crystalline structure

Fe (iron) crystals have a body-centered cubic (bcc) crystalline structure. Being given the molar mass of  Fe, $M =55.84 kg/kmol$, its mass density $\rho =7.8*10^3 kg/m^3$ and Avogadro number $N_A$ determine:

a)  The number of atoms $N_a$ from the unit (elementary) cell.

b) The lattice constant (distance between two consecutive parallel plans).

c) The number of atoms $n_a$ from a volume $V =1 cm^3$

d) The number of unit cells from a Fe crystal having a mass of $m=1 kg$

a) The number of atoms of a bcc unit cell is

$N_a =1+8*(1/8) =2 atoms$

b) The lattice constant $a_0$ is equal to the side of the cube representing the unit cell.  Therefore

$\rho*a_0^3 =N_a*(M/N_A)$

$a_0 =\left ( \frac{MN_a}{\rho*N_A} \right )^\frac{1}{3} =5.559*10^{-10} m$

c) the mass of volume $V=1 cm^3$ is

$m =\rho*V$

Total number of moles in volume V is

$n = m/M$

Total number of atoms in volume V is

$n_a =n*N_A =(\rho*V*N_A)/M =8.4*10^{12}$ atoms

d) volume of one unit cell is

$V_0 =a_0^3$

volume of mass crystal having $m =1 kg$ is

$V =m/\rho$

number of unit cells from mass $m= 1 kg$ is

$N_a =V/V_0 = m/(\rho*a_0^3) =7.184*10^{23}$ unit cells