Question

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 $300K=15$ mm)

• A. $16.45$ %
• B. $17.45$ %
• C. $14.45$ %
• D. $15.45$ %

Answer 1

The correct option is A $16.45$ %

From 715mm pressure, subtract aqueous tension 15mm to obtain pressure of nitrogen.
$P=715-15=700$ mm Hg
Volume of nitrogen at STP $=\frac{V\times P\times 273}{\left(T\right)\times 760}$

Volume of nitrogen at STP $=\frac{55\times 700\times 273}{\left(300\right)\times 760}=46$ mL
Percent of nitrogen $=\frac{volof{N}_{2}atSTP}{wtoforganiccompound}\times \frac{28}{22400}\times 100$
Percent of nitrogen $=\frac{46}{0.35}\times \frac{28}{22400}\times 100=16.45$ %
The percentage composition of nitrogen in the compound would be $16.45$ %.