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

Wt. of organic substance = 0.25 g 
$$\begin{gathered} V_1=40 mL,{ T}_1=300 K \\ {P}_1=725-25=700mm\text{ of }Hg \\ {P}_2=760mm\text{ of }Hg\left(\text{ at }STP\right) \\ {T}_2=273K \\ \frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2} \\ {V}_2\text{ (Volume of nitrogen at }STP\text{ ) } \\ =\frac{273\times 700\times 40}{300\times 760}=33.52mL \\ Percentage of nitrogen \\ =\frac{28\times \text{ volume of }{N}_2\text{ at }STP\times 100}{22400\times \text{ wt. of organic substance }} \\ =\frac{28\times 33.52\times 100}{22400\times 0.25}=16.76\% \end{gathered}$$