Question

In Duma's method for estimation of nitrogen, 0.25 g of an organic compound gave 40 mL of nitrogen collected at 300 K temperature and 725 mm pressure. If the aqueous tension at 300 K is 25 mm, the percentage of nitrogen in the compound is

• A. 18.20
• B. 16.76
• C. 14.76
• D. 20.36

Answer 1

Answer: (B)

Volume of nitrogen collected at 300 K and 725 mm pressure is 40 mL actual pressure

$=725-25=700$ $mm$

Using $\frac{\overset{\underset{c o n d i t i o n s}{G i v e n}}{P_{1} V_{1}}}{T_{1}}=\frac{\overset{A t S T P}{P_{2} V_{2}}}{T_{2}}$

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

Volume of nitrogen at STP $\left(\right.V_{2}\left.\right)=\frac{273 \times 700 \times 40}{300 \times 760}=33.52mL$

As 22,400 mL of $N_{2}$ at STP weight $=28g$

So, 33.5 mL of nitrogen weight $=\frac{28 \times 33.52}{22400}g$

N% $=\frac{28 \times V_{2} \times 100}{22400 \times W}$

Percentage of nitrogen (N%) $=\frac{28 \times 33.52 \times 100}{22400 \times 0.25}=16.76\%$