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CMC Medical CMC-Medical Ludhiana Solved Paper-2007

  • question_answer
    One mole of an anhydrous salt AB dissolves in water with the evolution of 21.0 J \[mo{{l}^{-1}}\]of heat. If the heat of hydration of AB is -29.4 J \[mo{{l}^{-1}}\], then the heat of dissociation of hydrated salt AB is

    A)  \[50.4\,J\,mo{{l}^{-1}}\]                              

    B)  \[8.4\,J\,mo{{l}^{-1}}\]

    C)  \[-50.4\,J\,mo{{l}^{-1}}\]            

    D)  \[-8.4\,J\,mo{{l}^{-1}}\]

    Correct Answer: B

    Solution :

                      (i) \[AB(s)+aq\xrightarrow{{}}AB(aq),\] \[\Delta H=-21\,J\,mo{{l}^{-1}}\] (ii) \[AB(s)+x{{H}_{2}}O\xrightarrow{{}}AB\cdot x\,{{H}_{2}}O(s),\] \[\Delta H=-29.4\,J\,mo{{l}^{-1}}\] Aim: \[AB\cdot x{{H}_{2}}O(s)+aq\xrightarrow{{}}AB(aq),\] \[\Delta H=?\] Eq. (i) is equivalent to \[AB(s)+x{{H}_{2}}O\xrightarrow{{}}AB\cdot x{{H}_{2}}O(s),\] \[\Delta H=\Delta \,{{H}_{1}}\] \[AB\cdot x{{H}_{2}}O(s)+aq\xrightarrow{{}}AB(aq),\] \[\Delta H=\Delta {{H}_{2}}\] \[\Delta {{H}_{1}}+\Delta {{H}_{2}}=-\,21\]                 \[-29.4+\Delta {{H}_{2}}=-21\] or                            \[\Delta {{H}_{2}}=8.4\,J\,mo{{l}^{-1}}\]


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