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

m=0.25 g, ${\mathrm{V}}_1$​=40 ml, ${\mathrm{T}}_1$​=300 K, ${\mathrm{P}}_1$​=725 mm−25 mm=700 mm.
 

${\mathrm{P}}_0$​=760 mm, ${\mathrm{T}}_0$​=273 K, ${\mathrm{V}}_0$​= ?

${\mathrm{V}}_0=\frac{{\mathrm{P}}_1{\mathrm{V}}_1}{{\mathrm{T}}_1}\times \frac{{\mathrm{T}}_0}{{\mathrm{P}}_0}=\frac{700\times 40}{300}\times \frac{273}{760}=33.53 \mathrm{mL}$

At STP, 22400 mL of nitrogen occupies 22400 $\mathrm{mL}$

22400 mL of ${\mathrm{N}}_2$​ at STP weighs = 28 gm
Hence the mass of nitrogen which corresponds to 33.53 mL is 22400 mL is $\frac{28\times 33.53}{224000}=0.0419 \mathrm{g}$

The percentage of nitrogen is $0.0419\times \frac{100}{0.25}=16.76 \%$