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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 12

  • question_answer
    Mixture of \[N{{H}_{3}}(g)\] and \[{{N}_{2}}{{H}_{4}}(s)\] are heated at \[1200K,\] as given below: \[{{P}_{i}}=0.3\,atm\]              \[{{P}_{f}}=2.7\,atm\] \[{{T}_{i}}=300\,K,\]                \[{{T}_{f}}=1200\,K\] \[{{V}_{i}}=VL\]                      \[{{V}_{f}}=VL\] The \[mol\,%\]of \[N{{H}_{3}}\]in the original mixture is

    A) \[25%\]                

    B)   \[20%\]                

    C) \[75%\]     

    D)    \[37.5%\]

    Correct Answer: C

    Solution :

      [c] Using equation \[PV=nRT\]
    Initial moles       Total final moles         
    \[{{n}_{1}}=\] moles of \[N{{H}_{3}}\] Moles formed from \[N{{H}_{3}}=\left( \frac{1}{2}+\frac{3}{2} \right){{n}_{1}}\]
    \[{{n}_{2}}=\] moles of \[{{N}_{2}}{{H}_{4}}\] Moles formed from \[{{N}_{2}}{{H}_{4}}=(1+2){{n}_{2}}=3{{n}_{2}}\]
    \[0.3\times V=({{n}_{1}}+{{n}_{2}})\times R\times 300\] \[2.7\times V=(2{{n}_{1}}+3{{n}_{2}})\times R\times 1200\] \[\frac{2{{n}_{1}}+3{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}=\left( \frac{9}{4} \right)\] \[\frac{{{n}_{1}}}{{{n}_{2}}}=3\] \[\frac{{{n}_{1}}}{{{n}_{2}}}=75%\]        


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