In a Kjeldahl's method for estimation of nitrogen, $\mathrm{NH}_{3}$ evolved from 0.6 g of the compo

Question

In a Kjeldahl's method for estimation of nitrogen, $\mathrm{NH}_{3}$ evolved from 0.6 g of the compound neutralized 15 mL of $0.5 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}$. The percentage of nitrogen in the compound is\

Professor PC Thomas Classes · Answer

15 mL of $0.5 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}$ $$ \equiv 30 \mathrm{~mL} 	ext { of } 0.5 \mathrm{M} \mathrm{NH}_{3} $$ Percentage of nitrogen $$ \begin{aligned} & =\frac{14 \mathrm{VN} 	imes 100}{1000 \mathrm{w}}=\frac{1.4 \mathrm{VN}}{\mathrm{w}} \ & =\frac{1.4 	imes 30 	imes 0.5}{0.6}=35 \% \end{aligned} $$