2. Sample Papers
  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main Sample Paper-35

  • question_answer
    The ammonia gas evolved from the treatment of \[0.30\] g of an organic compound for the estimation of \[{{N}_{2}}\], was passed in 100 ml of \[0.1\] M \[{{H}_{2}}S{{O}_{4}}\]. The excess of acid required 20 ml \[0.5\] NaOH, solution for complete neutralization. The organic compound is:

    A)  Acetamide        

    B)  Benzamide

    C)  Urea                                    

    D)  Thiourea

    Correct Answer: C

    Solution :

    \[{{H}_{2}}S{{O}_{4}}\,+2\,NaOH\,\xrightarrow{\,}N{{a}_{2}}S{{O}_{4}}\,+2{{H}_{2}}O\]             \[{{M}_{1}}\times n\,-factor\,\times {{V}_{1}}={{M}_{2}}\times n-factor\,\times {{V}_{2}}\]                         \[({{H}_{2}}S{{O}_{4}})\]             (NaOH)             \[0.1\times 2\times {{V}_{1}}=0.5\,\times 1\times 20\]             \[{{V}_{1}}=50\,ml\] Volume of \[{{H}_{2}}S{{O}_{4}}\] used up to absorb \[N{{H}_{3}}=50\,ml\]             \[N%\,=\frac{2.8\times \,(MV)}{W}\,=\frac{2.8\times 0.1\,\times 50}{0.3}\,=46.6\] Thus, the organic compound will be urea, which has 46.6 % \[{{N}_{2}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner