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NEET Chemistry The Solid State / ठोस प्रावस्था Question Bank Self Evaluation Test - The Solid State

  • question_answer
    \[CsBr\] crystallises in a body centered cubic lattice.  The unit cell length is 436.6 pm. Given that the atomic mass of Cs=133 and that of Br=80 amu and Avogadro number being \[6.02\times {{10}^{23}}mo{{l}^{-1}}\]  the density of \[CsBr\] is

    A) \[0.425\text{ }g/c{{m}^{3}}\]              

    B) \[8.5g/c{{m}^{3}}\]

    C) \[4.25\text{ }g/c{{m}^{3}}\]                

    D) \[82.5\text{ }g/c{{m}^{3}}\]

    Correct Answer: B

    Solution :

    [b] For body centred cubic lattice Z = 2 Atomic mass of unit cell \[=133+80=213\] a.m.u Volume of cell \[={{\left( 436.6\times {{10}^{-10}} \right)}^{3}}c{{m}^{3}}\] Density, \[\rho =\frac{ZM}{{{a}^{3}}{{N}_{A}}}\] \[=\frac{2\times 213}{{{\left( 436.6\times {{10}^{-10}} \right)}^{3}}\times 6.02\times {{10}^{23}}}\] \[=8.50g/c{{m}^{3}}\]


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