### Organic Chemistry Some Basic Principles and Techniques - Result Question 45

####51. 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 :

[2015]

(a) 18.20

(b) 16.76

(c) 15.76

(d) 17.36

## Show Answer

#### Solution:

- (b) Wt. of organic substance $=0.25 g$

$V_1=40 mL, T_1=300 K$

$P_1=725-25=700 mm$ of $Hg$

$P_2=760 mm$ of $Hg$ (at STP)

$T_2=273 K$

$\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}$

$V_2$ (Volume of nitrogen at STP)

$ =\frac{273 \times 700 \times 40}{300 \times 760}=33.52 mL $

Percentage of nitrogen

$ \begin{aligned} & =\frac{28 \times \text{ volume of } N_2 \text{ at STP } \times 100}{22400 \times \text{ wt. of organic substance }} \\ & =\frac{28 \times 33.52 \times 100}{22400 \times 0.25}=16.76 % \end{aligned} $