Manipal Engineering Manipal Engineering Solved Paper-2014

  • question_answer
    KBr crystallises in NaCI type of crystal lattice and its density is \[{{\sin }^{-1}}a+{{\sin }^{-1}}b+{{\sin }^{-1}}c=\pi ,\] Number of unit  3 cells of KBr present in a 1.00 mm3 grain of KBr are

    A) \[a.\sqrt{(1-{{a}^{2}})}+b.\sqrt{(1-{{b}^{2}})}+c.\sqrt{(1-{{c}^{2}})}\]

    B) \[\frac{1}{2}abc\]

    C) \[\frac{1}{3}abc\]                            

    D) \[\lambda \]

    Correct Answer: A

    Solution :

    In a unit cell of KBr, there are four formula units of KBr therefore, Density \[{{E}^{0}}({{O}_{2}}(g)/{{H}_{2}}O/O{{H}^{-}})=+0.4V]\] \[2Fe(s)+{{O}_{2}}(g)+2{{H}_{2}}O(l)\rightleftharpoons 2F{{e}^{2+}}(aq)\] \[+4O{{H}^{-}}(aq)\] Number of unit cells per \[E_{cell}^{0}=-0.48V\] \[E_{cell}^{0}=-0.04V\]                                


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