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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2015

  • question_answer
    In a closed container, a certain amount of \[\left[ -\frac{\sqrt{3}}{2},0 \right]\] is maintained at\[0{}^\circ C\]. At\[273{}^\circ C,\] the \[\left[ -\frac{\sqrt{3}}{2},0 \right)\cup \left( 0,\frac{\sqrt{3}}{2} \right]\]is completely dissociated to \[\left[ -\frac{\sqrt{3}}{2},\infty  \right)\] molecules. What will be its pressure as compared to initial pressure?

    A) Double                

    B) Three times

    C) Four times        

    D) Same

    Correct Answer: C

    Solution :

    \[\therefore \] \[\text{NaCl}=\frac{3.33}{4.44}\times \text{1}00=\text{75}%\] \[pH=p{{K}_{a}}+\log \frac{[salt]}{[acid]}\]                 \[5.0=4.0+\log \frac{[salt]}{acid}\] \[\frac{[salt]}{acid}=anti\log 1=10\]                         One mole of \[{{N}_{2}}{{O}_{4}}\]decomposes to gives 2 moles of\[=\frac{12}{44}\times \frac{0.66}{m}\times 100=\frac{18}{m}\]. \[=\frac{2}{18}\times \frac{0.36}{m}\times 100=\frac{4}{m}\]     \[={{C}_{3}}{{H}_{8}}\] \[\pi \] \[{{M}_{2}}=\frac{1000{{K}_{t}}\times {{W}_{2}}}{{{W}_{1}}\times \Delta T}=\frac{1000\times 5\times 2.5}{25\times 2.25}=222\]                 \[\therefore \]  \[i=\frac{{{M}_{T}}}{{{M}_{0}}}=\frac{122}{222}=0.5492=0.55\] \[\alpha =\frac{1-i}{1-\frac{1}{x}}-=\frac{1-0.55}{1-\frac{1}{2}}=\frac{0.45}{0.5}=0.9\] = four times the initial pressure.


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