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DUMET Medical DUMET Medical Solved Paper-2011

  • question_answer
    \[\alpha -(D)\]glucose\[\beta -(D)\]glucose, equilibrium constant for this is 1.8. The percentage of \[\alpha \]- glucose at equilibrium is

    A)  35.7            

    B)  55.6

    C)  44.4            

    D)  64.3

    Correct Answer: A

    Solution :

    \[\alpha \]- glucose \[\beta -(D)\] glucose
    1 0 : Initial cone.
    \[1-x\] \[x\] : Cone. at
    equilibrium
    \[\because \] \[{{K}_{eq}}=1.8\] \[\therefore \] \[\frac{x}{1-x}=1.8\] or \[x=1.8-1.8x\] or \[2.8x=1.8\] or \[x=\frac{1.8}{2.8}=0.643\] The concentration of \[\alpha \]- glucose \[=1-x\] \[=1-0.643=0.357\] and % concentration of \[\alpha \]- glucose \[=0.357\times 100\] \[=35.7%\]


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