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MGIMS WARDHA MGIMS WARDHA Solved Paper-2013

  • question_answer
    The heat of formation of\[{{C}_{12}}{{H}_{22}}{{O}_{11}}(s),C{{O}_{2}}(g)\] and\[{{H}_{2}}O(l)\]are\[-530,-94.3\]and\[-68.3\]kcal \[mo{{l}^{-1}}\]respectively. The amount of\[{{C}_{12}}{{H}_{22}}{{O}_{11}}\]to supply 2700 kcal of energy is

    A)  382.70 g                              

    B)  832.74 g

    C)  463.9 g                                

    D)  684.0 g

    Correct Answer: D

    Solution :

                    \[{{C}_{12}}{{H}_{22}}{{O}_{11}}(s)+12{{O}_{2}}(g)\xrightarrow[{}]{{}}12C{{O}_{2}}(g)+11{{H}_{2}}O(l)\] \[\Delta H_{comb}^{o}=[12\Delta H_{f}^{o}(C{{O}_{2}})]+11\Delta H_{f}^{o}({{H}_{2}}O]\] \[-\Delta H_{f}^{o}({{C}_{12}}{{H}_{22}}{{O}_{11}})]\] \[=-1352.9\text{ }kcal\] Thus, number of moles of\[{{C}_{12}}{{H}_{22}}{{O}_{11}}\] for getting 2700 kcal of heat                 \[=\frac{2700}{1352.9}=2\,mol\] \[=2\times 242=684g\]


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