Organic Chemistry Some Basic Principles and Techniques - Result Question 52
####58. 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 $300 K=15 mm$ )
[2011]
(a) 15.45
(b) 16.45
(c) 17.45
(d) 14.45
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Solution:
- (b) Given wt of compound taken $(w)=0.35 g$
Volume of nitrogen collected $(V)=55 mL$
Room temperature $(t K)=300 K$
Atmospheric pressure $(P)=715 mm$
Aq. tension $(\rho)=15 mm$
Calculation -
Volume of $N_2$ at NTP $=\frac{(P-\rho) \times V}{t} \times \frac{273}{760} mL$
$=\frac{(715-15) \times 55}{300} \times \frac{273}{760}=46.098 mL$
$%$ of nitrogen
$=\frac{28 \times \text{ vol. of } N_2 \text{ at } NTP \times 100}{22400 \times \text{ wt of organic compound }}$
$=\frac{28 \times 46.098 \times 100}{22400 \times 0.35}=16.46 %$
