Question

iron(II) oxide has a cubic structure and edge lenght of 5 angstorm. if density of oxide is 4gm cm-3, calculate the no. of Fe 2+ and O 2- present in each unit cell. [FeO = 72 of mol] N A = 6.022* 10 23 mol -1.

Answer 1

Answer:

The no. of $Fe^{2+}$ and $O^{2-}$ present in each unit cell is 4.

Explanation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density  of cubic structure

Z = number of atoms in unit cell

M = molecular mass of iron(II) oxide = 71.85 g/mol

$(N_{A})$ = Avogadro's number  

a = edge length of unit cell =$5\AA=5\times 10^{-8} cm$

$4g/cm^3=\frac{Z\times 71.85 g/mol}{6.022\times 10^{23} mol^{-1}\times (5\times 10^{-8} cm)^3}$

Z = 4.190 ≈ 4

The no. of $Fe^{2+}$ and $O^{2-}$ present in each unit cell is 4.