The Solid State

Question 161
CBSEENCH12005509

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
 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
Question 162
CBSEENCH12005510

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

 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

Question 163
CBSEENCH12005511

An element crystallizes in a structure having a fcc unit cell of an edge 200 pm. Calculate its density if 200 g of this element contains 24 x 1023 atoms.

Solution

Solution:
We have given that
Edge length of the unit cell
= 200 pm = 200 x 10–10 cm

Vol. of the unit cell
= (200 x 10–10 cm)= 8 x 10–24 cm3
Therefore, volume of the substance
= Vol. of unit cell x Vol. of 1 unit cell.

Since the element has a fcc unit cell, number of atoms per unit cell = 4.
Total number of atoms = Atoms/unit cell x number of unit cells.
                                     = 24 x 1023atoms = 4

atoms/unit cell x No. of unit cells.

∴   No. of unit cells = 24×1023 atoms4 atoms/unit cell

                               = 6×1023 unit cells.

∴  Volume of the substance
                   6×1023 unit cell × 8× 10-24 cm3/unit cell = 4.8 cm3


Now,     Density  = MassVolume               = 200g4.8 cm3 =4.17 g cm-3.



Question 164
CBSEENCH12005512

An element X with an atomic mass of 60 g/mol has density of 6.23 g/cm–3. If the edge length of its cubic unit cell is 400 pm, identify the type of cubic unit cell. Calculate the radius of an atom of this element.

Solution
Solution:
we have given that 
Density, d = 6.23 g/cm3
              a = 400 pm
              M = 60g/mol-1
 
Volume  = (a3)=(400)3 = (400×10-10 cm)3

Z=d×a3×NAM

NAM=6.023×1023 mol-160 g/mol 

Z = 6.023 g cm-3×(400)3×10-30 cm3×6.023×1023 mol-160 g mol-1    = 6.023×64×6.023600   = 2401.49600=4.

Hence, the type of cubic unit cell is FCC


Radius = aa2=40022=2002

              = 2001.414=141.4 pm. 

So radius of element is 141.4 pm

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