RAJASTHAN ­ PET Rajasthan PET Solved Paper-2012

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

    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\[=108\text{ }g\text{ }mo{{l}^{-1}}\] 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\[=\frac{1}{13.5}\times 108=8g\] Number of hydrogen atoms \[=\frac{8}{1}=8\] Weight of nitrogen\[=\frac{3.5}{13.5}\times 108=28g\] Number of nitrogen atom \[=\frac{28}{14}=2\] Hence, molecular formula \[={{C}_{6}}{{H}_{8}}{{N}_{2}}\]


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