### 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

## Show Answer

#### 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

$=\frac{28 \times \text{ vol. of } N_2 \text{ at } NTP \times 100}{22400 \times \text{ wt of organic compound }}$

$=\frac{28 \times 46.098 \times 100}{22400 \times 0.35}=16.46 %$