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 [2015]

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Solution

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 [2015]

1 16.76

2 15.76

3 17.36

4 18.20

Ans.
(1)

Mass of organic compound = 0.25 g

Experimental values, At STP

$V_{1} = 40 mL, V_{2} = ?, T_{1} = 300 K, T_{2} = 273 K$

$P_{1} = 725 - 25 = 700 mm, P_{2} = 760 mm$

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} \Rightarrow V_{2} = \frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}} = \frac{700 \times 40 \times 273}{300 \times 760} = 33.52$ $mL$

22400 mL of $N_{2}$ at STP weighs = 28 g

$∴$ $33.52 mL {of N}_{2} at STP weighs = \frac{28 \times 33.52}{22400} = 0.0419$ $g$

$% of N = \frac{Mass of nitrogen at STP}{Mass of organic compound taken} \times 100$

$= \frac{0.0419}{0.25} \times 100 = 16.76 %$

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