Question

In Duman'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 $300K=15$ mm)

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

Answer 1

The correct option is B $16.45$

In Duma's method of estimation of nitrogen:-
Calculation:- Volume of $N_2$ at NTP (By gas equation)
$\left(\frac{\rho -{\rho }_1}{t+273}\right)v\times \frac{273}{760}=V$ ml.
$\%$ of nitrogen in the given compound:
$\frac{28}{22400}\times \frac{V}{W}\times 100$
Here, $W=0.35$ gm.
$\rho =715$ mm (Pressure at which $N_2$ collected)
${\rho }_1=$ aqueous tension of water $=15$ mm.
$(t+273)K=300K$
$v$ ml = volume of moist nitrogen in nitrometer $=55$ ml.
So, volume of $N_2$ at NTP$=\left(V\right)=\frac{(715-15)\times 55}{300}\times \frac{273}{760}=46.098$ ml.
$\%$ of nitrogen $=\frac{28}{22400}\times \frac{46.098}{0.35}\times 100=16.45\%$.