The image above represents normally occupied positions.

To compute for normally occupied positions, four essential parameters are needed and these parameters are **Avogadro’s number (N _{A}), Density (ρ), Atomic weight (A_{K})** and

**Atomic weights (A**

_{G}).**The formula for calculating normally o**ccupied positions:

N = ^{NAρ }/ _{(AK + AG)}

N = Normally Occupied Positions

N_{A} = Avogadro’s Number

A_{K}, A_{G} = Atomic Weights

ρ = Density

Let’s solve an example;

Find the normally occupied positions when the avogadro’s number is 6.022e+23, atomic weight is 12, atomic weight is 16 and the density is 10.

This implies that;

N_{A} = Avogadro’s Number = 6.022e+23

A_{K}, A_{G} = Atomic Weights = 12, 16

ρ = Density = 10

N = ^{NAρ }/ _{(AK + AG)}

N = ^{(6.022e+23)(10)}/_{(12 + 16)}

N = ^{(6.022e+24)}/_{(28)}

N = 2.15e+23

Therefore, the **normally occupied positions **is **2.15e+23.**

**Calculating the Density when the Normally Occupied Positions, the Avogadro’s Number, the Atomic Weights is Given.**

ρ = ^{N (AK + AG)} / _{N}_{A}

Where;

ρ = Density

N = Normally Occupied Positions

N_{A} = Avogadro’s Number

A_{K}, A_{G} = Atomic Weights

Let’s solve an example;

Given that normally occupied positions is 20, the avogadro’s number is 6.022e+23, atomic weights is 18, 6.

This implies that;

N = Normally Occupied Positions =20

N_{A} = Avogadro’s Number = 6.022e+23

A_{K}, A_{G} = Atomic Weights = 18, 6

ρ = ^{N (AK + AG)} / _{N}_{A}

ρ = ^{20 (18 + 6)} / _{6.022e+23}

ρ = ^{20 (24)} / _{6.022e+23}

ρ = ^{480} / _{6.022e+23}

ρ = 7.970e-22

Therefore, the **density **is **7.970e-22.**

Continue reading How to Calculate and Solve for Normally Occupied Positions | Ceramics