Q13 / 60 · KVPY 2014 (SA) · Chemistry

The amonia evolved from $2 \mathrm{~g}$ of a compound in Kjeldahl's estimation of nitrogen neutralizes $10 \mathrm{~mL}$ of $2 \mathrm{M}$ $\mathrm{H}_{2} \mathrm{SO}_{4}$ solution. The weight percentage of nitrogen in the compound is

  • Correct answer
    A
    28
  • B
    14
  • C
    56
  • D
    7

Solution

$\begin{array}{c} 2 \mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \\ 10 \mathrm{ml} \\ 2 \mathrm{M} \end{array}$ millimole of $\mathrm{H}_{2} \mathrm{SO}_{4}=\frac{\mathrm{mmol} \text { of } \mathrm{NH}_{3}}{2}=20$ $\mathrm{mmol} \mathrm{NH}_{3}=\mathrm{mmol}$ of $N=40$ $W_{N}=\frac{40 \times 14}{1000}=\frac{560}{1000}=0.56 \mathrm{~g}$ $\%$ of $N=\frac{0.56}{2} \times 100=28$

Hence, the correct answer is option: 'A'.

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