WB JEE Medical WB JEE Medical 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) \[40\,g\,mo{{l}^{-1}}\]      

    B)  \[30\,g\,mo{{l}^{-1}}\]

    C) \[27\,g\,mo{{l}^{-1}}\]      

    D)  \[20\,g\,mo{{l}^{-1}}\]

    Correct Answer: C

    Solution :

     Given, cell is fee, so Z = 4 Edge length, \[a=404\,pm\,=4.04\,\times {{10}^{-8}}cm\] 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, \[d=\frac{Z\times M}{{{a}^{3}}.{{N}_{A}}}\] \[\therefore \] \[M=\frac{d.{{a}^{3}}.{{N}_{A}}}{Z}\] \[=\frac{2.72\times {{(4.04\,\times \,{{10}^{-8}})}^{3}}\times 6.02\times {{10}^{23}}}{4}\] \[=27\,g\,mo{{l}^{-1}}\]


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