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JEE Main & Advanced AIEEE Solved Paper-2005

  • question_answer
    An organic compound having molecular mass 60 is found to contain C = 20%, H = 6.67% and N = 46.67% while rest is oxygen. On heating it gives\[N{{H}_{3}}\]along with a solid residue. The solid residue give violet colour with alkaline copper sulphate solution. The compound is     AIEEE  Solved  Paper-2005

    A) \[C{{H}_{3}}C{{H}_{2}}CON{{H}_{2}}\]  

    B)        \[{{(N{{H}_{2}})}_{2}}CO\]          

    C)        \[C{{H}_{3}}CON{{H}_{2}}\]     

    D)        \[C{{H}_{3}}NCO\]

    Correct Answer: B

    Solution :

    Element Percentage \[\frac{\mathbf{Percentage}}{\mathbf{at}\mathbf{.wt}}\] Simple Ratio
    C 20.0 \[\frac{20.0}{12}=1.66\] \[\frac{1.66}{1.66}=1\]
    H 6.67 \[\frac{6.67}{1}=6.67\] \[\frac{6.67}{1.66}=4\]
    N \[46.67\] \[\frac{46.67}{14}=3.33\] \[\frac{3.33}{1.66}=2\]
    O 26.66 \[\frac{26.66}{16}=1.66\] \[\frac{1.66}{1.66}=1\]
    Empirical formula\[=C{{H}_{4}}{{N}_{2}}O\] Empirical formula weight \[=12+(4\times 1)+(2\times 14)+16=60\] \[\therefore \]\[n=\frac{Mol.\text{ }formula\text{ }weight}{Emp.formula\text{ }weight}=\frac{60}{60}=1\] \[\therefore \]Molecular formula\[=C{{H}_{4}}{{N}_{2}}O\] Given compound gives biuret test. Thus, given compound is urea\[{{(N{{H}_{2}})}_{2}}CO\] \[N{{H}_{2}}CON{{H}_{2}}+HNHCON{{H}_{2}}\xrightarrow{\Delta }\] \[\underset{biuret}{\mathop{N{{H}_{2}}CONHCON{{H}_{2}}}}\,+N{{H}_{3}}\xrightarrow{CuS{{O}_{4}}}Violet\text{ }colour\]


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