In Duma’s method for 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 $$25\,mm,$$  the percentage of nitrogen in the compound is

Solution :
Mass of the substance taken $$ = 0.25\,g$$
Volume of nitrogen collected $$ = 40\,ml$$
Atmospheric pressure $$ = 725\,mm$$
Room temperature $$ = 300\,k$$
Aqueous tension at $$300k = 25\,mm$$
Actual pressure of the gas $$ = \left( {725 - 25} \right)mmHg$$
                                           $$ = 700\,mm$$
To convert the volume at experimental conditions to volume at $$STP.$$
$$\eqalign{ & {\text{Experimental value }}\,\,\,\,\,\,\,\,\,\,{\text{At }}STP \cr & {P_1} = 700\,mm\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{P_2} = 760\,mm \cr & {V_1} = 40\,ml\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{V_2} = ? \cr & {T_1} = 300\,k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{T_2} = 273\,k \cr} $$
Substituting these values in the gase eq.
$$\frac{{{P_2}{V_2}}}{{{T_2}}} = \frac{{{P_1}{V_1}}}{{{T_1}}}$$
we get, $$\frac{{760 \times {V_2}}}{{273}} = \frac{{700 \times 40}}{{300}}$$
$${V_2} = \frac{{700 \times 40}}{{300}} \times \frac{{273}}{{760}} = 33.53\,ml$$
To convert volume at $$STP$$  into mass
$$22400\,ml$$   of nitrogen at $$STP$$  weigh $$= 28g$$
∴ $$33.53\,ml$$   of nitrogen at $$STP$$  will
$${\text{weigh}} = \frac{{28 \times 33.53}}{{22400 \times 0.25}}$$
To calculate percentage of nitrogen
$$\eqalign{ & = \frac{{28 \times 33.53}}{{22400 \times 0.25}} \times 100 \cr & = 16.76\% \cr} $$