KVPY Sample Paper KVPY Stream-SX Model Paper-29

  • question_answer
    Formation of polyethylene from calcium carbide takes place as follows
    \[n{{C}_{2}}{{H}_{4}}\xrightarrow{{}}{{\left( -C{{H}_{2}}-C{{H}_{2}}- \right)}_{n}}\]
    The amount of polyethylene obtained from 64.1 kg of \[Ca{{C}_{2}}\]is

    A) 7 kg

    B) 14 kg

    C) 21 kg

    D) 28 kg

    Correct Answer: D

    Solution :

    The concerned chemical reactions are
    (i) \[\underset{64kg}{\mathop{Ca{{C}_{2}}}}\,+2{{H}_{2}}O\to Ca{{\left( OH \right)}_{2}}+\underset{Ethyne,26kg}{\mathop{{{C}_{2}}{{H}_{2}}}}\,\]
    (ii) \[{{C}_{2}}{{H}_{2}}+{{H}_{2}}\to \underset{Ethylene,28kg}{\mathop{{{C}_{2}}{{H}_{4}}}}\,\]
    (iii) \[\underset{\begin{smallmatrix}  n\times 28kg \\  or\,28kg \end{smallmatrix}}{\mathop{n{{C}_{2}}{{H}_{4}}}}\,\to \underset{\begin{smallmatrix}  n\times 28kg\,polythene \\  or\,28kg \end{smallmatrix}}{\mathop{{{\left[ -C{{H}_{2}}-C{{H}_{2}}- \right]}_{n}}}}\,\]
    Thus 64 kg of \[{{\operatorname{CaC}}_{2}}\] gives 26 kg of acetylene which in turn gives 28 kg of ethylene whose
    28 kg gives 28 kg of the polymer, polythene.

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