The Solid State

Question 153

A metallic element exists as a cubic lattice. Each edge of the unit cell is 2.88 A°. The density of the metal is 7.20 g cm–3. How many unit cells will be there in 100 g of the metal?


we have given a= 2.88A0
Density = 7.20g cm-3
mass = 100g

Volume of the unit cell = (2.88 A°)3
                          = (2.88×10-8cm)3= 23.9 × 10-24 cm3

Volume of 100g of the metal

                 = WeightDensity=1007.20= 13.9 cm3

No. of unit cell in 13.9 cm3
                   = 13.9 cm323.9 × 10-24 cm3= 5.82 × 1023.

Question 154

The unit cell of an element of atomic mass 96, and density 10.3 g cm–3 is a cube with edge length of 314 pm. Find the structure of crystal lattice (simple cubic, F.C.C. or B.C.C.) Avogadro’s constant. NA = 6.023 x 1023 mol–1?


we have given

Density of element,  ρ=10.3 g cm-3
Cell edge  a= 314pm or 3.14 x 10-10 cm          

     NA =6.023×1023 mol-1

Atomic mass = 96 g mol-1

                 P = Z×Ma3×NAZ = P×a3×NAM

= 10.g cm-3×(3.14)3×10-30cm3×6.023×1023 mol-196 g mol-1=2

The structure of the crystal lattice is B.C.C.

Question 155

An element a crystallises in fcc structure. 200 g of this element has 4.12 x 1024atoms. The density of A is 7.2 g cm-3. Calculate the edge length of the unit cell?

we have given that
mass = 200g
density = 7.2 g cm-3
NA = 4.12 x 1024
Z = 4 
by using the following equation,
ρ = z× Ma3× NA

               a3 = Z×Md×N

         = 4×2007.2×4.12×1024= 80029.664×10-24

            a3 = 26.968×10-24a = 2.997×10-8a = 299.7 pm.
hence the edge length is 299.7pm
Question 156

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?


we have given that  
density, d= 7.2g cm-3
edge = 288 x 10-10 cm 
NA = 6.02 x 1023
for Bcc structure Z= 2

Atomic mass = ?


d = ZMa3NAM = d×a3×NAZM=7.25×(2.88×10-10)×6.02×10232g/molM= 51.77 g/mol of atoms
Thus, atomic mass of element is =51.77 g/mol

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