JEE Main & Advanced AIEEE Solved Paper-2002

  • question_answer
    In an organic compound of molar mass \[108\,g\,mo{{l}^{-1}}\] C, H and N atoms are present in \[9:1:3.5\] by weight. Molecular formula can be   AIEEE  Solved  Paper-2002

    A) \[{{C}_{6}}{{H}_{8}}{{N}_{2}}\]  

    B)                           \[{{C}_{7}}{{H}_{10}}N\]  

    C)           \[{{C}_{5}}{{H}_{6}}{{N}_{3}}\]     

    D)           \[{{C}_{4}}{{H}_{18}}{{N}_{3}}\]  

    Correct Answer: A

    Solution :

    Molar mass 108g    Total part by weight \[=9+1+3.5=13.5\]                    Weight of carbon \[=\frac{9}{13.5}\times 108=72\,\,g\]    Number of carbon atoms \[=\frac{72}{12}=6\]                    Weight of hydrogen \[=\frac{1}{13.5}\times 108=8\,\,g\] Number of hydrogen atoms \[=\frac{8}{1}=8\] Weight of nitrogen \[=\frac{3.5}{13.5}\times 108=28\,g\]              Number of nitrogen atom \[=\frac{28}{14}=2\] Hence, molecular formula \[={{C}_{6}}{{H}_{8}}{{N}_{2}}\]


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