0.2033 g of an organic compound in the Duma’s method gave 31.7 mL of nitrogen at ${14}^{0}C$ and 758 mm pressure. The percentage of nitrogen in the compound is (Aqueous tension at ${14}^{0}C$= 14mm)

$\text{\%\hspace{0.17em}\hspace{0.17em}N\hspace{0.17em}=\hspace{0.17em}}\frac{{\text{mass\hspace{0.17em}\hspace{0.17em}of\hspace{0.17em}N}}_{\text{2}}}{\text{22400}}\text{×\hspace{0.17em}}\frac{\text{Vol\hspace{0.17em}.\hspace{0.17em}×100\hspace{0.17em}}}{\text{mass\hspace{0.17em}\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}compound}}$

${\mathrm{V}}_1=31.7 {\mathrm{P}}_1=760-\text{ tension }$

= 746

$T_{1 }= 287K$

$P_2=760,T_2=273,V_2=?$

$\frac{{\text{P}}_{\text{1}}{\text{\hspace{0.17em}V}}_{\text{1}}}{{\text{T}}_{\text{1}}}\text{=}\frac{{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{T}}_{\text{2}}}$

${\text{V}}_{\text{2}}\text{\hspace{0.17em}=\hspace{0.17em}}\frac{\text{746\hspace{0.17em}\hspace{0.17em}×\hspace{0.17em}\hspace{0.17em}31.7\hspace{0.17em}\hspace{0.17em}×\hspace{0.17em}273}}{\text{760\hspace{0.17em}\hspace{0.17em}×\hspace{0.17em}\hspace{0.17em}287}}\text{\hspace{0.17em}=\hspace{0.17em}29.5}$

$\text{\%\hspace{0.17em}\hspace{0.17em}N}=\frac{28}{22400}\times \frac{29.5\times 100}{0.2033}$

= 18.2