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

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

Tardigrade Admin · Accepted Answer

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

= 725 - 25 = 700

mm

Using

\frac{P _{1} V _{1} co n d i t i o n s G i v e n}{T _{1}} = \frac{P _{2} V _{2} A tSTP}{T _{2}}

V_{2} = \frac{P _{1} V _{1} \times T _{2}}{T _{1} \times P _{2}}

Volume of nitrogen at STP

(V_{2}) = \frac{273 \times 700 \times 40}{300 \times 760} = 33.52 m L

As 22,400 mL of

N_{2}

at STP weight

= 28 g

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%

Q. 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

18.20

16.76

14.76

20.36