In Duma's method for estimation of nitrogen, 0.25 g of an organic compound gave 40mL of nitrogen collected at 300 K temperature and 725 mm pressure. If the aqueous tension at 300K is 25 mm, the percentage of nitrogen in the compound is

Mass of organic compound = 0.25 g

$$\begin{gathered} \mathrm{Experimental} \mathrm{values}, \mathrm{At} \mathrm{STP} \\ {\mathrm{V}}_1=40 \mathrm{mL} {\mathrm{V}}_2=? \\ {\mathrm{T}}_1=300 \mathrm{K} {\mathrm{T}}_2=273 \mathrm{K} \\ {\mathrm{P}}_1=725-25=700 \mathrm{mm} {\mathrm{P}}_2=760 \mathrm{mm} \\ \frac{{\mathrm{P}}_1{\mathrm{V}}_1}{{\mathrm{T}}_1}=\frac{{\mathrm{P}}_2{\mathrm{V}}_2}{{\mathrm{T}}_2} \\ {\mathrm{V}}_2=\frac{{\mathrm{P}}_1{\mathrm{V}}_1{\mathrm{T}}_2}{{\mathrm{T}}_1{\mathrm{P}}_2}=\frac{700\times 40\times 273}{300\times 760}=33.52 \mathrm{mL} \\ 22400 \mathrm{mL}\text{ of }{\mathrm{N}}_2\text{ at }\mathrm{STP}\text{ weighs }=28 \mathrm{g} \\ \therefore 33.52 \mathrm{mL}\text{ of }{\mathrm{N}}_2\text{ at STP weighs }=\frac{28\times 33.52}{22400} \\ =0.0419 \mathrm{g} \\ \%\text{ of }\mathrm{N}=\frac{\text{ Mass of nitrogen at STP }}{\text{ Mass of organic compound taken }}\times 100 \\ =\frac{0.0419}{0.25}\times 100=16.76\% \end{gathered}$$