The Solid State

  • Question 249
    CBSEENCH12010071

    An element with molar mass 27 g mol-1  forms a cubic unit cell with edge length 4.05 x 10-8 cm . If its density is 2.7 g cm-3 , what is the nature of the cubic unit cell?

    Solution

    Molar mass of the given element, M = 27 g mol-1 = 0.027 kg mol-1

    Edge length, a = 4.05 x 10-8 cm = 4.05 x 10-10 m

    Density, d = 2.7 g cm-3 = 2.7 x 103 kg m-3

     Applying the relation,

     straight d space equals fraction numerator straight z space straight x space straight m over denominator straight a cubed space straight x space straight N subscript straight A end fraction

    Where, Z is the number of atoms in the unit cell and NA is the Avogadro number. Thus,

     

    
straight Z equals space fraction numerator straight d space straight x space straight a cubed space straight x space straight N subscript straight A space over denominator straight m end fraction

fraction numerator 2.7 space straight x space 10 cubed space straight x space left parenthesis 4.05 space straight x space 10 to the power of negative 10 end exponent right parenthesis cubed space straight x space 6.022 space straight x space 10 to the power of 23 over denominator 0.027 end fraction space equals 4

    Since the number of atoms in the unit cell is four, the given cubic unit cell has a face-centred cubic (fcc) or cubic-closed packed (ccp) structure.

    Question 250
    CBSEENCH12010094

    An element with density 11.2 g cm-3  forms a f.c.c. lattice with edge length of 4 x10-8

    Calculate the atomic mass of the element. (Given:  NA = 6.022x 10-23 (mol-1

    Solution

    Density, d = 11.2 g cm-3

    Edge length, a = 4x10-8 cm

    Avogadro number, NA = 6.022x1023 mol-1

    Number of atoms present per unit cell, Z (fcc) = 4


    We know for a crystal system,

     straight d equals space fraction numerator straight z space straight x space straight m over denominator straight a cubed space straight x space straight N subscript straight A end fraction

straight m space equals fraction numerator straight d space straight x space straight a cubed space straight x space straight N subscript straight A over denominator straight Z end fraction

We space get comma

straight M space equals fraction numerator 11.2 space straight x space 64 space straight x space 10 to the power of negative 24 end exponent space straight x space 6.022 straight x space 10 to the power of 23 over denominator 4 end fraction equals space 107.91 space straight g

       

    Thus, the atomic mass of the element is 107.91 g.

     

    Question 251
    CBSEENCH12010095

    Examine the given defective crystal:
     

    Answer the following questions:


    (i) What type of stoichiometric defect is shown by the crystal?

    (ii) How is the density of the crystal affected by this defect?

     (iii) What type of ionic substances shows such defect? 

    Solution

    (i) Schottky defect is shown by the mentioned crystal, as an equal number of cations and anions are missing in the crystal lattice.

    (ii) This defect leads to decrease in density, as an equal number of the cations and anions are missing from the crystal lattice. A number of such defects in ionic solids are quite significant. For example, in NaCl, there are approximately 106 Schottky pairs per cm3 at room temperature.

    (iii) This kind of defect is shown by that ionic substance in which the cations and anions are of almost similar sizes. 


    Examples:  NaCl, KCl and CsCl.  

    Question 252
    CBSEENCH12010137

    How many atoms constitute one unit cell of a face-centered cubic crystal? 

    Solution

    Number of atoms in one face centred cubic unit cell can be determined:-

     

    (i) 8 corners atoms × 1/8 per corner atom =  1 over 8 x 8= 1 atom

    (ii) 6 face-centered atoms × 1/ 2 atom per unit cell = 1 half x 6= 3 atoms

     

    ∴ Total number of atoms per unit cell = 4 atoms

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