A metal (at. mass = 50) has a bcc crystal structure. The density of the metal is 5.96 g cm–3. Find the volume of its unit cell?

 Solution:
We have given that
Atomic mass of the metal = 50g
bcc unit cell, Z = 2
Density of metal = 5.96 g/cm
3
Therefore, volume of the unit,


a3 = M×ZNA×d     = 50×26.02×1023×5.96    = 2.787 × 10-23 cm3
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An atom has fcc crystal whose density is 10 gm–3 and cell edge is 100 pm. How many atoms are present in its 100 g?

Solution:
We have given that
Density = 10 gm-3
Mass = 100g
edge of unit cell ,a= 4 since it is a Fcc crystal

we have to find total number of atom, So by following relation we can get the result,

Total No. of atoms = Z×Ma3.d

Therefore, number of atoms

=4×100 g(100×10-12m)3×10 gm-3= 4×1031 atoms
 
thus the number of atoms is 4 x 1031 

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An element crystallises in BCC structure. The edge length of its unit cell is 288 pm. If the density of crystal is 7.2 g cm–3, what is the atomic mass of the element?

For bcc structure
        
For bcc structure        

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Calculate the density of silver which crystallizes in a face centred cubic lattice with unit cell length 0.4086 nm (At. mass of Ag = 107.88)

 Solution:
We have given that

Unit cell length,
a = 0.40806 nm  = 4.086 x 10-10 m

  If fcc lattice the number of atoms per unit cell,
i.e. Z = 4

        M for Ag = 107.88 g mol-1

                      =1.0788×10-1 kg mol-1

              NA = 6.023 × 1023


Density of Ag, d = ZMNA a3

                         = 4×1.0788×10-16.023×1023×(4.086×10-10)3= 1.051 × 104 kg m-3

 Thus the density of silver is 1.051 x 104 kgm-3

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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:
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.

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