The Solid State

  • Question 149
    CBSEENCH12005497

    If NaCl is doped with 10–3 mol % of SrCl2' what is the concentration of cation vacancies?

    Solution

    We know that two Na+ ions are replaced by each of the Sr2+ ions while SrCl2 is doped with NaCl. But in this case, only one lattice point is occupied by each of the Sr2+ ions and produce one cation vacancy.

    Here 10 – 3 mole of SrCl2 is doped with 100 moles of NaCl Thus, cation vacancies produced by NaCl = 10 – 3 mol Since, 100 moles of NaCl produces cation vacancies after doping = 10 –3 mol

    Therefore, 1 mole of NaCl will produce cation vacancies after doping
    10 to the power of negative 3 end exponent over 100 space equals 10 to the power of negative 5 end exponent space mol
    therefore, total cationic vacancies
    =10-5 x Avogadro's number
    =10-5  x 6.023 x 1023
    =6.023 x 10-18 vacancies
     

    Question 150
    CBSEENCH12005498

    The ionic radius of CI ion is 181 pm. Consider the closest packed structure in which all anions are just touching:
    (i) Calculate the radius of the cation that just fits into the octahedral holes of this lattice of anions. (ii) Calculate the radius of the cation that just fits into the tetrahedral holes of this lattice of anions.

    Solution
    given:
    Radius = 181 pm
    thus,
    Radius of octahedral void = 0.414 r = 0.414 x 181 pm = 74.934 pm
    Cation having radius 74.934 pm will just fit into octahedral voids.

    Radius of tetrahedral void

    = 0.225 r = 0.225 x 181 pm = 40.725 pm
    Cation having radius 40.725 pm will just fit into tetrahedral void.
    Question 151
    CBSEENCH12005499

    An element occurs in BCC structure with cell edge of 300 pm. The density of the element is 5.2 g cm–3. How many atoms of the element does 200 g of the element contain?

    Solution
    solution:
    we have given that
    z = 2 
    a = 300 pm
    d = 5.2 gm cm
    –3
    by using the given formula we obtaine mass of element i.e.           
    Question 152
    CBSEENCH12005500

    Iron (II) oxide has a cubic structure and each unit cell has side 5A. If the density of the oxide is 4 g cm–3, calculate the number of Fe2+ and O2– ions present in each unit cell. (Molar mass of FeO = 72 g mol–1, NA = 6.02 x 1023 mol–1).

    Solution

    Solution:
    We  have given
    volume of the unit cell = (5 x 10-8cm)3 = 1.25 x 10-22cm3

    Density of FeO = 4g cm-3

    Density,

    ρ = z×Ma3×NA4= z×72(5×10-8)3×6.02×1023z= 4×1.25×10-22 ×6.02×102372 = 4.18 4

    Each unit cell has four units of FeO. So it has four Fe2+ and four O2– ions.

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