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JEE Main & Advanced JEE Main Paper (Held On 10 April 2015)

  • question_answer
    1.4 g of an organic compound was digested according to Kjeldahl's method and the ammonia evolved was absorbed in 60 mL of \[M/10\,{{H}_{2}}S{{O}_{4}}\] solution. The excess sulphuric acid required 20 mL of M/10 NaOH solution tor neutralization. The percentage of nitrogen in the compound is: JEE Main Online Paper (Held On 10 April 2015)

    A)  24

    B)  5

    C)  10

    D)  3

    Correct Answer: C

    Solution :

                     Mili equivalents of \[{{H}_{2}}S{{O}_{4}}=60\times \frac{M\times 2}{10}=12\] Mili equivalents of \[NaOH=20\times \frac{M}{10}=2\] Mili equivalents of \[N{{H}_{3}}=12-2=10\] % of Nitrogen \[=\frac{1.4\times \left( N\times V \right)N{{H}_{3}}}{W}\] \[=\frac{1.4\times 10}{1.4}=10\]


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