The Solid State

  • Question 153
    CBSEENCH12005501

    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?

    Solution

    solution: 
    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
    CBSEENCH12005502

    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?

    Solution

    solution:
    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
    CBSEENCH12005503

    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?

    Solution
    solution:
    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
    CBSEENCH12005504

    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?

    Solution

    Solution:
    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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