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AMU Medical AMU Solved Paper-2012

  • question_answer
    Ammonia undergoes self-dissociation according to the reaction,\[2N{{H}_{3}}(l)N{{H}_{4}}^{+}(am)+NH_{2}^{-}(am)\]where am, stands for ammoniated. When 1 mole of \[N{{H}_{4}}Cl\] is dissolved in 1 kg of liquid ammonia, the b. p. at 760 tour is observed at \[-{{32}^{o}}C\] (Normal b.p. of \[N{{H}_{3}}(l)\] is\[-{{33.4}^{o}}C\]). What conclusions are reached about the nature of the solution?

    A)  \[N{{H}_{4}}Cl\] is completely dissociated in \[N{{H}_{3}}\]

    B)  \[N{{H}_{4}}Cl\] is partially dissociated in \[N{{H}_{3}}\]

    C)  \[N{{H}_{4}}Cl\] is not dissociated in \[N{{H}_{3}}\]

    D)  Boiling point is not raised

    Correct Answer: A

    Solution :

                    Vapour pressurs, \[\pi \propto bp\,(T)\] \[\therefore \]                  \[\frac{{{\pi }_{1}}}{{{\pi }_{2}}}=\frac{{{T}_{1}}}{{{T}_{2}}}\] \[\Rightarrow \]                               \[\frac{760}{{{\pi }_{2}}}=\frac{240.3\,K}{306.4\,K}\]                                 \[{{\pi }_{2}}=\frac{760\times 306.4}{240.3}=969\,torr\] \[\because {{\pi }_{2}}>{{\pi }_{1}}\] that means more gases are present or \[N{{H}_{4}}Cl\] gets completely dissociated into ammonia gas.


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