Question

A crystal of A₂0 is Na₂O type solid with density of 1000 kg m-3. What is the nearest distance between two A+ ions in A? (Molar mass of A₂0 is 76.8 g mol™ 1, N₁ = 6 x 102³)​

Answer 1

Given - Crystal of $A_2O$ is Na₂O type solid with a density of $1000 kg m^{-3}.$

Find - What is the nearest distance between two A+ ions in A?

Solution-$d=\frac{z\times M\omega}{a^3\times N_a}$

$Z=4$

$a=(\frac{4\times 76.8}{1000 \times 6 \times 10^{23}} )^{1/3}$

$a=(\frac{512}{10^{27}} )^{1/3}$

$a=8\omega m$

$x=4\sqrt{3}\omega m$

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Answer 2

Concept:

Density formula,

ρ  = (Z × M)/(N₁ × a³)

where ρ is density, z is the number of atoms in the unit cell, M is the molar mass, a is the edge of the unit cell, and N₁ is Avagadro's number.

Given:

Crystal of Na₂O type solid, Crystal is A₂O.

Density, ρ = 1000 kg m⁻³

Molar mass of A₂O, M = 76.8 g/mol =  76.8 × 10⁻³ kg/mol

Avagadro's number, N₁ = 6 × 10²³

Find:

The nearest distance between the two ions A⁺ ions and A⁻.

Solution:

If the edge length of the unit cell is a, then the nearest distance between the two ions A⁺ ions and A⁻ is a/√2.

Na₂O-type solids are having face-centered cubic structures. So, Z = 4.

Substituting the values in the density formula and calculating edge length,

ρ  = (Z × M)/(N₁ × a³)

1000 kg m⁻³ = (4 × 76.8 × 10⁻³ kg/mol )/(6 × 10²³ mol⁻¹ × a³)

a³ = (4 × 76.8 × 10⁻³ )/(6 × 10²³ × 1000)

a³ = 51.2 × 10⁻²⁹ m³

a³ = 512 × 10⁻³⁰ m³

a = ∛(512 × 10⁻³⁰ m³ )

a = 8 × 10⁻¹⁰ m

Distance between the ions = (8 × 10⁻¹⁰ )/√2 =  4√2× 10⁻¹⁰ m.

Hence, the nearest distance between the two ions A⁺ ions and A⁻ is  4√2× 10⁻¹⁰ m.

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