# The Solid State

Question 157

## Caesium chloride crystallises as a body centred cubic lattice and has a density of 4.0 g cm–3.  Calculate the length of the edge of the unit cell of caesium chloride crystal.         [Molar Mass of CsCl = 168.5 g mol–1, NA = 6.02 x 1023 mol–1]

Solution

Solution;
we have given that

or

Question 158

## The density of chromium is 7.2 g cm–3. If the unit cell is cubic with edge length of 289 pm, determine the type of the unit cell (Atomic mass of Cr = 52 amu).

Solution

Solution:
We have given that

Gram atomic mass of Cr(M) = 52.0 g mol-1
Edge length of unit cell (a) = 289 pm
Density of unit cell $\left(\mathrm{\rho }\right)$ = 7.2 g cm-3

Avogadro's Number

or              $\mathrm{Z}=\frac{\mathrm{\rho }×{\mathrm{a}}^{3}×{\mathrm{N}}_{\mathrm{A}}×{10}^{-30}}{\mathrm{M}}$

Since the unit cell has 2 atoms, it is body centre in nature.

Question 159

## Iron (II) oxide has a cubic structure and each of the unit cell is 5.0 A°. If density of the oxide is 4.0 g cm-3, calculate the number of Fe2+ and O2– ions present in each unit cell.

Solution

Volume of the unit cell = (5 A°)
= (5 x 10–8 cm)3
= 125 x 10–24 cm3
= 1.25 x 10–22 cm3
Density of FeO = 4.0 g cm–3

Therefore, mass of the unit cell
= 1.25 x 10–22 cm3 x 4 g cm–3
= 5 x 10–22g
Mass of all molecules of FeO

Number of FeO molecules/unit cell

Thus, there are 4 Fe
2+ ions and 4O2– ions in each unit cell.

Question 160

## An element (At. mass 60) have face centred cubic structure has a density of 6.23 g cm–3. What is the edge length of the unit cell?

Solution
Solution:
we have given that
Atomic mass of element = 60.
Number of atoms per fcc unit cell = 4
Density of the element = 6.23 g/cm
3

Density,         $\mathrm{d}=\frac{\mathrm{N}×\mathrm{M}}{{\mathrm{N}}_{\mathrm{A}}×{\mathrm{a}}^{3}}$

or

Therefore, edge of unit cell, a = 400 pm.