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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2007

  • question_answer
        The ammonia evolved from the treatment of 0.30g of an organic compound for the estimation of nitrogen was passed in 10s) ml, of M sulphuric acid. The excess of acid required 20 mL of 0.5 M sodium hydroxide solution for complete neutralization. The organic compound is

    A)  acetamide         

    B)  benzAmide

    C)  urea                                     

    D)  thiourea

    Correct Answer: C

    Solution :

                    Let unreacted \[0.1\,M(=0.2\,N){{H}_{2}}S{{O}_{4}}=V\,mL\] \[\therefore \] 20 mL of\[0.5M\,NaOH\] \[=VmL\,of\,0.2\,N\,{{H}_{2}}S{{O}_{4}}\] \[\therefore \]  \[20\times 0.5=V\times 0.2\] \[\therefore \]  \[V=50mL\] Used \[{{H}_{2}}S{{O}_{4}}=100-50=50\text{ }mL\] % Nitrogen\[=\frac{1.4NV}{w}.\] Where  \[N=\]normality of \[{{H}_{2}}S{{O}_{4}}\] \[V=\]volume of\[{{H}_{2}}S{{O}_{4}}\]used \[\therefore \]  %Nitrogen \[=\frac{1.4\times 0.2\times 50}{0.30}\] \[=46.67%\] % of nitrogen in (a) \[C{{H}_{3}}CON{{H}_{2}}=\frac{14\times 100}{59}=23.73%\]  (b)\[{{C}_{6}}{{H}_{5}}CON{{H}_{2}}=\frac{14\times 100}{122}=11.48%\]  (c)\[N{{H}_{2}}CON{{H}_{2}}=\frac{28\times 100}{60}=46.67%\]  (d)\[N{{H}_{2}}CSN{{H}_{2}}=\frac{28\times 100}{76}=36.84%\]


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