VMMC VMMC Medical Solved Paper-2012

  • question_answer
    The number of unit cells in the 5.85g crystals of NaCI are

    A) \[1.5\times {{10}^{23}}\]

    B)  \[1.5\times {{10}^{22}}\]

    C)  \[3.0\times {{10}^{22}}\]

    D)  \[3.0\times {{10}^{23}}\]

    Correct Answer: C

    Solution :

     Density of unit cell \[(d)=\frac{Z\times M}{{{a}^{3}}.{{N}_{A}}}\] Mass of a unit cell, \[(m)=\] volume \[\times \] density \[={{a}^{3}}\times \frac{Z\times M}{{{a}^{3}}.{{N}_{A}}}\] \[=\frac{Z\times M}{{{N}_{A}}}=\frac{4\times 58.5}{6.023\times {{10}^{23}}}\] \[=3.885\times {{10}^{-22}}\] Therefore, number of unit cells in 5.85g NaCl crystals \[=\frac{5.85}{3.885\times {{10}^{-22}}}=1.5\times {{10}^{22}}\]


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