Question

In duma’s method $0.52g$ of an organic compound on combustion gave $68.6$ mL $N_{2}$ at $27^{o}C$ 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)

$14.93\%$

Answer 2

: $V_{1} = 68.6mL,P_{1} = 756mm,T_{1} = 300K$

$V_{2} = ?,P_{2} = 760mm,T_{2} = 273K$

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

At NTP , vol. of

$N_{2},V_{2} = \frac{P_{1}V_{1}}{T_{1}}.\frac{T_{2}}{P_{2}} = \frac{756 \times 68.6}{300} \times \frac{273}{760}$

$= 62.09mL$

Percentage of nitrogen in organic compound

$= \frac{28}{22400} \times \frac{V_{2}}{w} \times 100 = \frac{28}{22400} \times \frac{62.09}{0.52} \times 10 = 14.93\%$