NEET Chemistry NEET PYQ-The Solid State

  • question_answer
    A metal has a fee lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g \[c{{m}^{-3}}\]. The molar mass of the metal is            (\[{{N}_{A}}\]Avogadro's constant \[=6.02\times {{10}^{23}}mo{{l}^{-1}}\])                                       [NEET 2013]

    A)                   \[40\,g\,mo{{l}^{-1}}\]

    B) \[30\,g\,mo{{l}^{-1}}\]

    C) \[27\,g\,mo{{l}^{-1}}\]

    D) \[20\,g\,mo{{l}^{-1}}\]

    Correct Answer: C

    Solution :

    [c] Given, cell is fee, so Z = 4
    Edge length, a = 404 pm = 4.04 x 10-8 cm
    Density of metal, \[d=2.72\,g\,c{{m}^{-3}}\]
    \[{{N}_{A}}=6.02\times {{10}^{23}}mo{{l}^{-1}}\]
    Molar mass of the metal, M = ?
    We know that density, \[d=\frac{Z\times M}{{{a}^{3}}\cdot {{N}_{A}}}\]
    \[\therefore \]\[M=\frac{d\cdot {{a}^{3}}\cdot {{N}_{A}}}{Z}\]
    \[=\frac{2.72\times {{(4.04\times {{10}^{-8}})}^{3}}\times 6.02\times {{10}^{23}}}{4}\]
    \[=27\,g\,mo{{l}^{-1}}\]


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