\(\rho=\frac{\mathrm{n} \times \mathrm{M}}{\mathrm{a}^{3} \mathrm{~N}_{\mathrm{A}}}\)
For bcc n =2, M = 39 nearest distance 2r = 4.52 A⁰
\(a =\frac{4r}{\sqrt {3}} = \frac { 2\times 4.52 \times 10^{-10} }{ \sqrt{3}} \)
\(\rho =\frac { {( 5.21\times 10^{-10})^{ 3 }} \times (6.023 \times 10^{-23}) }{ 2 \times 39} \)
ρ = 915 kgm^-3

If a is the length of the side, then the length of the leading diagonal passing through the body centered atom is \(\sqrt{3a} \)
Required distance =  \(\left( \cfrac { \sqrt { 3 } }{ 2 } \right) a\)

sc ⇒ 2r = a
⇒ r = a/2
bcc ⇒ 4r = \( \sqrt{3a} \) ⇒ r = \(\frac { \sqrt{3a}} {4}\)
fcc ⇒ 4r ⇒  \( \sqrt{2a} \) ⇒ r = \(\frac { \sqrt{2a}} {4} = \frac { a} {2\sqrt{2a}}\)
\(\left( \frac { a }{ 2 } :\frac { \sqrt { 3 } }{ 4 } a:\frac { a }{ 2\sqrt { 2 } } \right) \)

let edge length = a
\(\sqrt{2a} = 4r\)
\(a = \frac{4 \times 300}{\sqrt {2}}\)
a = 600 x 1.414
a = 848.4 pm

Packing efficiency = 68%
\(\therefore\)empty space percentage = 100 - 68 = 32%

For a fcc structure = r_x+ / r_y- = 0.414
Given that r_x⁺ = 100 pm
r_y = 100pm/0.414

\(3 \sqrt{a} = r_{C_{s+}} + 2r_{C_{f-}} + r_{C_{s+}} \)
\(\left( \frac { \sqrt { 3 } }{ 2 } \right) a = ​​ r_{C_{s+}} + r_{C_{f-}} \)
\(\left( \frac { \sqrt { 3 } }{ 2 } \right) \times 400\)= inter ionic distance

\(\left( \frac { r_{c^+} }{ r_{A^-} } \right) \) = \(\left( \frac { 0.98 \times 10^{-10 }}{ 1.81 \times 10^{-10} } \right) = 0.54\)
It is in the range of 0.414 to 0.732, hence the coordination number of each ion is 6.

If the total number of M atoms is n, then the number of tetrahedral voids = 2n
Given that \(\left( \frac { 1 }{ 3 } \right) ^{rd}\) of tetrahedral voids are occupied.
i.,e \(\left( \frac { 1 }{ 3 } \right) \times 2n\) are occupied by N atoms
\(\therefore\)M : N = n : \(\left( \frac { 2 }{ 3 } \right) n\)
= 1 : \(\frac { 2 }{ 3 }\)
Hence M₃N₂ = 3 : 2

In bcc unit cell,
2 atoms = 1 unit cell
Number of atoms in 8 g of element is, Number of moles = 8g / 40 g mol^-1 = 0.2 mol
1 mole contains 6.023 x10²³ atoms 
0.2 mole contains 0.2 x 6.023 x 10²³ atoms
[1 unit cell / 2 atoms] x 0.2 x 6.023 x 10²³
= 6.023 x 10²² unit cells.

Lattice points are occupied by CO₂ molecules

If number of close packed atoms = N, then
The number of Tetrahedral holes formed = 2N
The number of Octahedral holes formed = N
Therefore, N:2N = 1:2

Number of A ions = N_c/8 = 8/8 = 1
Number of B ions = N_f/2 = 6/2 = 3
Simplest formula = AB₃