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VMMC VMMC Medical Solved Paper-2008

  • question_answer
    Formaldehyde polymerizes to form glucose according to the reaction \[6HCHO{{C}_{6}}{{H}_{12}}{{O}_{6}}\] The theoretically computed equilibrium constant for this reaction is found to be \[6\times {{10}^{22}}.\] If 1 M solution of glucose dissociates according to the above equilibrium, the concentration of formaldehyde in the solution will be

    A) \[1.6\times {{10}^{-2}}M\]

    B) \[1.6\times {{10}^{-4}}M\]

    C) \[1.6\,\times {{10}^{-6}}M\]

    D) \[1.6\,\times {{10}^{-8}}\,M\]

    Correct Answer: B

    Solution :

    A very high value of K for the given equilibrium shows that dissociation of glucose to form HCHO is very-very small. Hence, at equilibrium, we can take, \[[{{C}_{6}}{{H}_{12}}{{O}_{6}}]=1\,M\]                 \[K=\frac{[{{C}_{6}}{{H}_{12}}{{O}_{6}}]}{{{[HCHO]}^{6}}}ie,6\times {{10}^{22}}=\frac{1}{{{[HCHO]}^{6}}}\] or \[[HCHO]={{\left( \frac{1}{6\times {{10}^{22}}} \right)}^{1/6}}=1.6\times {{10}^{-4}}M\]


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