AIIMS AIIMS Solved Paper-2015

  • question_answer
    The volume of a colloidal particle, \[C{{H}_{3}}-NH-NH-C{{H}_{3}}\,and\,{{C}_{2}}{{H}_{5}}OH\] as compared to volume of solute particle \[\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{N}{{\text{H}}_{\text{2}}}\text{ and }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH}\] in a true solution could be

    A) \[N{{a}^{+}}\]                                   

    B) \[C{{l}^{-}}\]

    C) \[{{\tan }^{-1}}\left[ \frac{{{p}^{2}}+PQ+{{Q}^{2}}}{PQ} \right]\]              

    D) \[{{\tan }^{-1}}\left[ \frac{{{p}^{2}}+{{Q}^{2}}-PQ}{PQ} \right]\]

    Correct Answer: B

    Solution :

    For true solution, the diameter range is 1 to \[\text{1}0\text{{ }\!\!\mathrm{\AA}\!\!\text{ }}\] and for colloidal solution, is 10 to\[\text{1}000\text{ { }\!\!\mathrm{\AA}\!\!\text{ }}\] \[\therefore \]  \[\frac{{{V}_{C}}}{{{V}_{s}}}=\frac{4/3\pi r_{c}^{3}}{4/3\pi r_{s}^{3}}={{\left( \frac{{{r}_{c}}}{{{r}_{s}}} \right)}^{3}}={{\left( \frac{10}{1} \right)}^{3}}={{10}^{3}}\]


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