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JEE Main & Advanced JEE Main Paper (Held On 11 April 2014)

  • question_answer
    A gaseous compound of nitrogen and hydrogen contains 12.5% (by mass) of hydrogen. The density of the compound relative to hydrogen is 16. The molecular formula of the compound is:   [JEE Main Online Paper ( Held On 11 Apirl  2014 )

    A) \[N{{H}_{2}}\]                  

    B) \[{{N}_{3}}H\]

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

    D) \[{{N}_{2}}{{H}_{4}}\]

    Correct Answer: D

    Solution :

                    In an unknown compounds containing N and H given % of H = 12.5% \[\therefore \]% of N = 100 ? 12.5 = 87.5%
    Element Percentage Atomic ratio Simple ratio
    H 12.5% \[\frac{12.5}{1}=12.5\] \[\frac{12.5}{6.25}=2\]
    N 87.5 \[\frac{87.5}{14}=6.25\] \[\frac{6.25}{6.25}=1\]
    \[2\times \]vapour density = Mol. wt = mol wt. \[=16\times 2=32.\] Molecular formula \[=n\times \] empirical formula mass \[n=\frac{32}{16}=2\] \[\therefore \]Molecular formula of the compound will be \[={{(N{{H}_{2}})}_{2}}\]\[={{N}_{2}}{{H}_{4}}\]


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