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VMMC VMMC Medical Solved Paper-2011

  • question_answer
    2g of mixture of CO and \[\text{C}{{\text{O}}_{\text{2}}}\]on reaction with excess \[{{\text{I}}_{\text{2}}}{{\text{O}}_{\text{5}}}\]produced 2.54 g of \[{{\text{I}}_{2}}\].What would be the mass % of \[\text{C}{{\text{O}}_{\text{2}}}\]in the original mixture?

    A) 60                                          

    B) 30                          

    C)  70                         

    D)         35

    Correct Answer: B

    Solution :

    \[5CO+{{I}_{2}}{{O}_{5}}\xrightarrow{{}}5C{{O}_{2}}+{{I}_{2}}\] 1 mole of \[{{\text{I}}_{\text{2}}}\text{=1}\] mole of \[{{\text{I}}_{\text{2}}}{{\text{O}}_{\text{5}}}=5\] moles of CO Hence, mole of \[CO=5\times \frac{2.54}{254}=0.05\] Mass of \[CO=0.05\times 28=1.4\,g\] Mass of \[C{{O}_{2}}=2-1.4=0.6g\] Mass \[%\,C{{O}_{2}}=\frac{0.6\times 100}{2}=30\]


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