NEET NEET SOLVED PAPER 2013

  • 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 (\[{{N}_{A}}\] Avogadros constant \[=6.02\times {{10}^{23}}mo{{l}^{-1}}\])

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

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

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

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

    Correct Answer: C

    Solution :

    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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