Question

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 $715\, mm$ pressure. The percentage composition of nitrogen in the compound would be
(Aqueous tension at $300 \,K = 15\, mm$ )

• 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)
$\, \, \, \, \, \, \Bigg(\frac{\rho-\rho_1}{t+273}\Bigg)V_1\times\frac{273}{760}=V_2$
Where, $p_1=\rho-\rho_1$
$\, \, \, \rho=715mm$ (pressure at which $N_2$ collected)
$\rho_1=$ aqueous tension of water =15 mm
$T_1=t+273=300 K$
$\, \, \, \, \, V_1=55 mL$ 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
$\, \, \, \, \, =\frac{28}{22400}\times\frac{V_2}{W}\times100=-\frac{28}{2400}\times\frac{46.098}{0.35}\times100$
$\, \, \, \, \, \, =16.45$ %