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Iron (II) oxide crystallise in cubic structure with unit cell edge of 5.0 Å If the density of the oxide is 3.8 g cm-3 . Calculate the no.of Fe2+ and O2- present in each unit cell, [atomic mass of Fe = 56, O = 16]

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Cell edge a = 5.0Å = 5.0 × 10 m = 5 × 10-8 cm

Density = 3.8 g/cm3 

Molecular mass of FeO = 56 + 16 = 72 u

NA = 6.022 × 1023

d = \(\frac{ZM}{N_Aa^3}\)

Z = \(\frac{dN_Aa^3}{M}\)

\(\frac{3.8\times6.022\times10^{23}}{72}\times\)(5 × 10-8)3

= 3.97 = 4

Each cell contain 4 FeO molecule every FeO molecule contain one Fe2+ and one O2- ion. 

So no. of Fe2+ ion =4 no.of O2- ion = 4.

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