In Duma’s method of estimation of nitrogen $$0.35\,g$$  of an organic compound gave $$55\,ml$$  of nitrogen collected at $$300\,K$$  temperature and $$175\,mm$$  pressure. The percentage composition of nitrogen in the compound would be ( Aqueous tension at $$300\,K = 15\,mm$$   )

Solution :
According to combined gas equation,
$$\frac{{{p_1}{V_1}}}{{{T_1}}} = \frac{{{p_2}{V_2}}}{{{T_2}}}$$
Where, $${p_2} = $$  pressure of $${N_2}$$  at $$STP$$  $$ = 760\,mm$$
$${T_2} = $$  Temperature of $${N_2}$$  at $$STP$$  $$ = 273\,K$$
$${V_2} = ?$$
Volume of $${N_2}$$  at $$STP$$  ( By gas equation )
$$\left( {\frac{{p - {p_1}}}{{t + 273}}} \right){V_1} \times \frac{{273}}{{760}} = {V_2}$$
Where, $${p_1} = p - {p_1}$$
$$p = 715\,mm$$   ( pressure at which $${N_2}$$  collected )
$${p_1} = $$  aqueous tension of water $$ = 15\,mm$$
$${T_1} = t + 273 = 300\,K$$
$${V_1} = 55mL = $$   volume of moist nitrogen in nitrometer
$$\eqalign{ & \therefore \,\,{V_2} = \frac{{\left( {715 - 15} \right) \times 55}}{{300}} \times \frac{{273}}{{760}} \cr & = 46.098\,mL \cr} $$
$$\% $$  of nitrogen in given compound
$$\eqalign{ & = \frac{{28}}{{22400}} \times \frac{{{V_2}}}{W} \times 100 \cr & = \frac{{28}}{{22400}} \times \frac{{46.098}}{{0.35}} \times 100 \cr & = 16.45\% \cr} $$