The Solid State

Question 157
CBSEENCH12005505

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    

   Z =1, ρ = 4 g cm-3,
        M = 168.5 mol-1,NA = 6.02×1023mol-1   ρ = Z×Ma3×N

or          a3 = Z×Mρ×NA    = 1×168.5g mol-14×6.02×1023    = 168.5×10-2324.08
log a3 = log 168.5 + log 10-23 - log 24.083 log a = 2.2266 - 23.000-1.38163 log a = -22.1550+2-23 log a = 24+(2.1550)       a = Antilog 8¯.7183 = 5.228×10-8cm         = 5.228×10-8×1010pm = 522.8 pm.

Question 158
CBSEENCH12005506

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 (ρ) = 7.2 g cm-3
  
 Avogadro's Number (N0) = 6.022×1023 mol-1       

     ρ = Z×Ma3×NA×10-30

or              Z=ρ×a3×NA×10-30M
   
   Z=(7.2 g cm-3) × (289)3× (6.022×1023 mol-3) × (10-30 cm3)(52.0 g mol-1)=2

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

Question 159
CBSEENCH12005507

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
=726.022×1023= 1.195 × 10-22 g

 Number of FeO molecules/unit cell
=5×10-22g1.195×10-22g= 4.194

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

Question 160
CBSEENCH12005508

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

                              
= 6.231023 g/(pm)3

Density,         d=N×MNA×a3


or              a3 = N×Md×NA      =60×4×10306.23×6.02×1023      = 64 × 106 (pm)3


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

NCERT Book Store

NCERT Sample Papers

Entrance Exams Preparation