Question

In Duma's method of estimation of nitrogen $0.35g$ of an organic compound gave $55mL$ of nitrogen collected at $300K$ temperature and $715mm$ pressure. The percentage composition of nitrogen in the compound would be
(Aqueous tension at $300K=15mm$ )

• A. 16.45
• B. 17.45
• C. 14.45
• D. 15.45

Answer 1

Answer: (A)

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)
$\left(\frac{\rho -{\rho }_1}{t+273}\right)V_1\times \frac{273}{760}=V_2$
Where, $p_1=\rho -{\rho }_1$
$\rho =715mm$ (pressure at which $N_2$ collected)
${\rho }_1=$ aqueous tension of water =15 mm
$T_1=t+273=300K$
$V_1=55mL$ volume of moist nitrogen in
nitrometer
$\therefore V_2=\frac{(715-15)\times 55}{300}\times \frac{273}{760}=46.098mL$
% of nitrogen in given compound
$=\frac{28}{22400}\times \frac{V_2}{W}\times 100=-\frac{28}{2400}\times \frac{46.098}{0.35}\times 100$
$=16.45$ %