The Solid State

  • Question 145
    CBSEENCH12005493

    Copper crystallizes into a fcc lattice with edge length 3.61 x 10–8 cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm–3.

    Solution

    We have given that
    Length of edge, a = 3.61x10–8 cm
    Atomic mass of cupper, M = 63.55 g/mol
    Avogadro constant, NA = 6.022x1023 per mol
    Number of atoms in unit cell of FCC, Z= 4
    Use formula of density,
    straight p equals fraction numerator ZM over denominator straight a cubed straight N subscript straight A end fraction

Putting space the space given space value space in space above space equation

straight p space equals fraction numerator 4 space straight x space 63.55 over denominator left parenthesis 3.61 space straight x 10 to the power of negative 8 end exponent right parenthesis cubed space left parenthesis 6.022 space straight x 10 to the power of 23 right parenthesis end fraction straight g over Cm cubed

straight p equals space fraction numerator 254.2 over denominator left parenthesis 47.04 space space straight x 10 to the power of negative 24 end exponent right parenthesis space left parenthesis 6.022 space straight x 10 to the power of 23 right parenthesis end fraction

straight p equals space fraction numerator 254.2 over denominator 28.318 end fraction straight g over cm cubed

straight p equals 8.92 space straight g divided by cm cubed

     



    Question 146
    CBSEENCH12005494

    Analysis shows that nickel oxide has formula Ni 0.98 O1.00 What fractions of Nickel exist as Ni2+ and Ni3+ ions? 

    Solution

    Formula is Ni0.98O1.00
    So the ration of Ni : O = 98:100
    So if there are 100 atoms of oxygen then 98 atoms of Ni,
    Let number of atoms of Ni+2 = x
    Then number of atoms of Ni+3 = 98–x
    Charge on Ni = charge on O
    So that oxygen has charge –2
    3(98–x) + 2x = 2 (100)
    294 –3x +2x = 200
    –x = – 94
    x = 94
    Percentage of Ni+2 = (atom of Ni+2/total number of atoms of Ni)100
    =100(94/98)x100
    = 96%
    Percentage of Ni+3
    =100 – Ni+2
    =100 – 96
    = 4 %


    Question 147
    CBSEENCH12005495

    Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the length of a side of the cell?

    Solution
    Radius, r = 0.144 nm,
    Unit cell is fcc, edge of unit cell, a = ?
    In face centered unit cell, diagonal of face = 4r = a2
                                    [∵ face diagonal = a2]

    Therefore,       a=4r2 = 2r
                         
      = 2 × 1.4142 × 0.144 nm= 0.4073 nm.

    Length of side of unit cell,
                       
    a = 0.4073 nm.
                       
     
    Question 148
    CBSEENCH12005496

    Aluminium crystallizes in a cubic close-packed structure. Its metallic radius is 125 pm.
    (a) What is the length of the side of the unit cell?
    (b) How many unit cells are there in 1.00 cm3 of aluminium?

    Solution

    Given, radius of atom (r) = 125 pm
    (a) For ccp structure, we know that,

    straight r equals space fraction numerator straight a over denominator 2 square root of 2 end fraction

where comma space straight r space equals space radius space and space straight a space equals space length space of space side space
equals space 125 space pm space equals space fraction numerator straight a over denominator 2 square root of 2 end fraction

straight a equals space 125 space straight x space 2 square root of 2

straight a equals 125 space straight x 2 space 1.414 space pm

straight a equals 353.5 space pm

left parenthesis straight b right parenthesis volume space of space 1 space unit space cell space equals straight a cubed

equals left parenthesis 353.5 space straight x space 10 to the power of negative 3 end exponent cm right parenthesis cubed
equals 44192902.36 space straight x 10 to the power of negative 30 end exponent space cm cubed
equals 442 space straight x space 10 to the power of negative 25 end exponent space cm cubed

Thus comma space number space of space unit space cell space of space aluminium space in space 1 space cm cubed

equals fraction numerator 1 cm cubed over denominator 442 space straight x space 10 to the power of negative 25 end exponent space cm cubed end fraction

equals 2.27 space straight x 10 to the power of 22 space unit space cell


     

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