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MGIMS WARDHA MGIMS WARDHA Solved Paper-2015

  • question_answer
    4.0 g molecular\[PC{{l}_{5}}\] were heated in a vessel of 6L volume. At equilibrium state, 50% of\[PC{{l}_{5}}\]is dissociated. The equilibrium constant tor the reaction is

    A)  \[0.66\,mol\,{{L}^{-1}}\]             

    B)  \[0.33\,mol{{L}^{-1}}\]

    C)  \[0.44\,mol\,{{L}^{-1}}\]             

    D)  \[0.88\,mol\,{{L}^{-1}}\]

    Correct Answer: B

    Solution :

                    \[PC{{l}_{5}}PC{{l}_{3}}+C{{l}_{2}}\]
    Moles before dissociation \[C\] 0 0
    Moles after dissociation \[C(1-\alpha )\] \[C\alpha \] \[C\alpha \]
    \[{{K}_{c}}=\frac{[PC{{l}_{3}}][C{{l}_{2}}]}{[PC{{l}_{5}}]}=\frac{{{C}^{2}}{{\alpha }^{2}}}{C(1-\alpha )}=\frac{C{{\alpha }^{2}}}{(1-\alpha )}\] Given,\[C=\frac{4}{6}\]and\[\alpha =0.5\]         \[\left[ \because C=\frac{mole}{litre} \right]\]                 \[{{K}_{C}}=\frac{4\times {{(0.5)}^{2}}}{6\times (1.05)}=0.33\,mol\,{{L}^{-1}}\]


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