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
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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} $
