Question

In Duma’s method of estimation of nitrogen, 0.55 g of an organic compound gave 60 mL of nitrogen collected at 300 K temperature and 715 mm pressure. The percentage composition of nitrogen in the compound would be:
(aq. tension at 300 K = 15 mm)

• A. $11.4\%$
• B. $4.6\%$
• C. $18.6\%$
• D. $25\%$

Answer 1

The correct option is A $11.4\%$

Volume of $N_2$ at STP
(By gas equation):

$\frac{p_1\times V_1}{T_1}=\frac{p_2\times V_2}{T_2}$
Here
$p_1=\rho -{\rho }_1\rho =715mm\left(\text{pressure at which}{N}_2\text{collected}\right)$
${\rho }_1=\text{aqueous tension of water}=15mm$

$V_2=\frac{(\rho -{\rho }_1)\times V_1}{T_1}\times \frac{273}{760}$
$V_1=60mL$ volume of moist nitrogen in nitro meter
$p_2=1atm=760mm{T}_2=0^{\circ }C=273K$

$V_2=\frac{(715-15)\times 60}{300}\times \frac{273}{760}=50.29mL$
$\%N_2=\frac{28}{22400}\times \frac{V_2}{W}\times 100$

W= Weight of organic compound.
$\%N_2=\frac{28}{22400}\times \frac{50.29}{0.55}\times 100$

$\%N_2=11.43\%$