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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2003

  • question_answer
    A compound having mol., \[mass=78\]contains \[C=92.31%\]and \[H=7.69%.\]Its molecular formula is

    A)  \[{{C}_{5}}{{H}_{12}}\]     

    B)  \[{{C}_{5}}{{H}_{18}}\]

    C)  \[{{C}_{4}}{{H}_{30}}\]       

    D)  \[{{C}_{6}}{{H}_{6}}\]

    Correct Answer: D

    Solution :

    The empirical formula of compound can be calculated as
    Element Percentage Percentage/ At.wt. Ratio
    \[C\] \[92.31\] \[~92.31/12=7.7~\] \[1\]
    \[H\] \[7.69\] \[7.69/1=7.69~\] \[1\]
    Thus the empirical formula is \[CH\] The empirical formula \[wt=12+1=13\] Value of n \[=\frac{Molecular\,wt}{Empirical\,formula\,wt}\]                 \[=\frac{78}{13}=6\] Molecular formula \[=(Empirical\text{ }formula)\times n\]                 \[=(CH)\times 6={{C}_{6}}{{H}_{6}}\]


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