Question

In Duma’s method $0.52g$ of an organic compound on combustion gave $68.6$ mL $N_2$ at $27^{\circ }C$ and $756$ mm pressure. What is the percentage of nitrogen in the compound?

• A. $12.22\%$
• B. $14.93\%$
• C. $15.84\%$
• D. $16.23\%$

Answer 1

Answer: (B)

$V_1=68.6mL$, $P_1=756mm$, $T_1=300K$
$V_2=?$, $p_2=760mm$, $T_2=273K$
$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$
At $NTP$, vol. of $N_2,V_2=\frac{P_1{V}_1}{T_1}\cdot \frac{T_2}{P_2}=\frac{756\times 68.6}{300}\times \frac{273}{760}$
$=62.09mL$
Percentage of nitrogen in organic compound
$=\frac{28}{22400}\times \frac{V_2}{w}\times 100$
$=\frac{28}{22400}\times \frac{62.09}{0.52}\times 100$
$=14.93\%$