The Solid State

  • Question 237
    CBSEENCH12005585

    An element crystallises in fcc structure with an edge of 200 pm. Calculate density if 200 g of this element contains 24 x 1023 atoms. 

    Solution

    we have 
    edge length =200Pm
    volume of the unit = ( 200 x10-10 cm)3
                             = 8 x10-24 cm-3

    In a FCC unit cell there are four atoms per unit cell.therefore  mass of unit cell  =200 x 424 x1023 =33.3 x10-23gDensity  =mass of unit cell volume of unit cell =33.3 x10-23g8 x10-24cm3  =41.6 g cm-3

    density = 41.6 g cm-3.





    Question 238
    CBSEENCH12005586

    The nearest neighbour Ag atoms in the silver crystal are 2.87 x 10–10 m apart. What is the density of silver? Silver crystallises in fcc form.

    Solution

    we have given :
    mass of silver = 107.87 g
    Z(FCC)          = 4
    distance between nearest neighbour Ag atoms = 2.87 x10-10

    face diagonal 2 =2.87 x10-10 m2a2 =2.87 x10-10 m a= 2.87 x10-10 x 2a= 2.87 x10-10 x 1.414a= 4.05 x 10-10now using formula d= Z x Ma3 x NAd =4 x 107.87(4.05 x 10)3 x 6.02 x1023d= 10.84 g cm-3

    Question 239
    CBSEENCH12005587

    The cell edge of a fcc element of atomic mass 108 is 409 pm. Calculate its density.

    Solution

    we have given,.
    mass =108 g
    edge (a) = 409pm  
    a3 = 6.84 x1023 cm3
    Na =6.023 x10-233
    Z =4

    Apply formula

     d= Z xMa3 x NAd = 4 x 1086.84 x 1023 x 6.02 x10-23 = 10.98 g cm-3

    density =  10.98 g cm–3

    Question 240
    CBSEENCH12005588

    An element ‘A’ of atomic mass 100 having bcc structure has unit cell edge of 400 pm. Calculate the density of ‘A’ and the number of unit cells for 10 g of ‘A’.

    Solution

    length of the unit cell edge = 400Pm =400x10-10 cm
     volume of the unit cell  = ( 400 x10-10cm)3
                                     = 6.4 x 10-23 cm-3

    as the element A forms a body centred  cubic lattice so no of atom per units cell is 2

    z= 2 atoms unit cell

    atomic mass of the element  =100 g/mol
    density of element is given by

    md= M xZa3 xNA d= 100 x26.4 x 10-23 x 6.023 x 10 23 = 51.88 g/cm3volume of 10 g of A = massDensity                             = 10g5.188g cm-3                           = 1.9275 cm3number of unit cell in 1.9275 cm3 volume  =volume of the substance unit cell volume                                                           = 3.0 x1022 unit cell



















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