Question

An element with molar mass 63 g / mol forms a cubic unit cell with edge length of 360.8 pm. If its density is 8.92 g/ cm3 . What is the nature of the cubic unit cell?

Answer 1

1. simple cubic = 1 atom/ unit crystal
2. body centred cubic = 2 atom/crystal
3. face centred cubic = 3 atom / crystal
4. hexagonal close packing = 4 atom / crystal

Answer 2

Answer : The nature of cubic unit cell is, FCC (face centered cubic unit cell) and the number of atom in unit cell is, 4

Solution : Given,

Density = $8.92g/cm^3$

Edge length = $360.8pm=360.8\times 10^{-10}cm$     $(1pm=10^{-10}cm)$

Atomic mass of an element(M) = 63 g/mole

Avogadro's number $(N_{A})=6.022\times 10^{23} mol^{-1}$

Formula used :  

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$      .............(1)

where,

$\rho$ = density

Z = number of atom in unit cell

M = atomic mass

$(N_{A})$ = Avogadro's number  

a = edge length of unit cell

Now put all the values in above formula (1), we get

$8.92g/cm^3=\frac{Z\times (63g/mol)}{(6.022\times 10^{23}mol^{-1}) \times(360.8\times 10^{-10}Cm)^3}$

$Z=4.0046=4$

The number of atoms in unit cell is, 4 that means the nature of unit cell is FCC (face centered cubic unit cell).

Therefore, the nature of cubic unit cell is, FCC (face centered cubic unit cell) and the number of atom in unit cell is, 4