Manipal Engineering Manipal Engineering Solved Paper-2014

  • question_answer
    For the reaction represented by the equation, \[\frac{496}{729}\]9.0 g of \[\frac{400}{729}\] completely reacts with 1.74 g of oxygen. The approximate molar mass of X will be

    A) 20                                          

    B) 40

    C) 60                                          

    D)  80

    Correct Answer: D

    Solution :

    \[\text{m}{{\text{s}}^{\text{-1}}}\] \[{{C}_{6}}{{H}_{5}}CH=N-N=CH{{C}_{6}}{{H}_{5}}\]        \[N{{H}_{3}}\] 9.0g        1.74g. \[N{{H}_{2}}N{{H}_{2}}\]reacts with \[-{{\left( C{{H}_{2}}-\underset{\begin{smallmatrix}  | \\  Cl \end{smallmatrix}}{\mathop{CH}}\, \right)}_{n}}-\] \[-{{\left( C{{H}_{2}}-\underset{\begin{smallmatrix}  | \\  CN \end{smallmatrix}}{\mathop{CH}}\, \right)}_{n}}-\] \[-{{\left( C{{H}_{2}}-\underset{\begin{smallmatrix}  | \\  Cl \end{smallmatrix}}{\overset{\begin{smallmatrix}  Cl \\  | \end{smallmatrix}}{\mathop{C}}}\, \right)}_{n}}-\] will react with \[\xrightarrow[{}]{{}}2\] Molecular mass \[C{{l}_{2}},\]or             331 = 12 + 4 XX \[C{{l}_{2}},FeC{{l}_{3}}\]


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