What will be the theoretical density of Lithium fluoride in g/cm³ LiF having rock salt NaCl structure. The ionic radii of lithium and fluoride is 0.068 nm and 0.133 nm ?

 

What will be the theoretical density of Lithium fluoride in g/cm³ LiF having rock salt NaCl structure. The ionic radii of lithium and fluoride is 0.068 nm and 0.133 nm ?

 Solution :-

Given,    Radius of lithium, `r_l`= 0.068 nm

              Radius of Fluoride,`r_f` = 0.133 nm

 

Now ,   we know that for rock salt structure

  Number of Lithium atoms in a unit cell,`n_l` = 4

Number of Fluoride atoms in a unit cell, `n_f`= 4

Also, molecular weight of Li ,`M_l` = 6.94 g/mol

Also, molecular weight of F, `M_f`= 19 g/mol

 

 

Also, each side of unit cell, a = 2(`r_l` + `r_f`) = 2(0.068 +0.133) = 0.402 nm = 4.02X`10^-8` cm

Also, the volume of unit cell will be, V = `a^3`=`{(4.02x10^{-8})}^3`

 

So, density of LiF is given by

Density, ρ =`\frac{n_l\times M_l+n_f\times M_f}{V\times N_A}`, where `N_A`is Avogadro number

             ρ = `\frac{4\times6.94+4\times19}{{(4.02\times10^{-8})}^3\times6.022\times10^{23}}`= 2.65 g/cm³

 

Hence, the theoretical density of LiF is 2.65 g/cm³