In a Dumas method for estimation of nitrogen, 0.3 g of an organic compound gave 50 mL of nitrogen co

Question

In a Dumas method for estimation of nitrogen, 0.3 g of an organic compound gave 50 mL of nitrogen collected at 300 K and 715 mm\ pressure (aqueous tension $=15 \mathrm{~mm}$ at 300 K ) .The percentage of nitrogen in the compound is\

Professor PC Thomas Classes · Answer

$\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}} \quad \frac{(\mathrm{P}-\mathrm{q}) \mathrm{V}}{\mathrm{T}}=\frac{\mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{~T}_{0}}$\ $\frac{(715-15) 50}{300}=\frac{760 	imes \mathrm{V}_{0}}{273} \quad 	herefore \mathrm{~V}_{0}=42 \mathrm{~mL}$\ Mass of $22400 \mathrm{~mL} \mathrm{~N}_{2}$ at STP $=28 \mathrm{~g}$\ $	herefore$ Mass of $42 \mathrm{~mL} \mathrm{~N}_{2}=\frac{28}{22400} 	imes 42 \mathrm{~g}$\ $\%$ of nitrogen $=\frac{	ext { mass of nitrogen }}{	ext { mass of org. comp. }} 	imes 100$ $$ =\frac{28}{22400} 	imes \frac{42}{0.3} 	imes 100=17.5 \% $$