2. Solved Papers
  3. Manipal Engineering

Manipal Engineering Manipal Engineering Solved Paper-2012

  • question_answer
    The vapour density of\[{{N}_{2}}{{O}_{4}}\]at a certain temperature is 30. What is the percentage dissociation of\[{{N}_{2}}{{O}_{4}}\]at this temperature?

    A) 46%                                       

    B) 53%

    C) 75%                       

    D)        92%

    Correct Answer: B

    Solution :

    The reaction is                 \[{{N}_{2}}{{O}_{4}}2N{{O}_{2}}\] Molecular weight of\[{{N}_{2}}{{O}_{4}}=2\times 14+4\times 16\]                                                    \[=92\] \[\therefore \]Vapour density\[(D)\]of\[{{N}_{2}}{{O}_{4}}\]                                                  \[=\frac{molecular\,\,weight}{2}\]                                                   \[=\frac{92}{2}=46\] From,                                \[\alpha =\frac{D-d}{(n-1)d}\]                                             \[\alpha =\frac{46-30}{(2-1)30}=\frac{16}{30}\]                                                  \[=0.533=53.3%\]


You need to login to perform this action.
You will be redirected in 3 sec spinner