In Duma's method for the 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 Hg, the percentage of nitrogen in the compound is :

A
18.20
B
16.76
C
15.75
D
17.36

Solution:

m=0.25 g, $V_{1}$ =40 ml, $T_{1}$ =300 K, $P_{1}$ =725 mm−25 mm=700 mm.

$P_{0}$ =760 mm, $T_{0}$ =273 K, $V_{0}$ = ?

$V_{0} = \frac{P_{1} V_{1}}{T_{1}} \times \frac{T_{0}}{P_{0}} = \frac{700 \times 40}{300} \times \frac{273}{760} = 33.53 mL$

At STP, 22400 mL of nitrogen occupies 22400 $mL$

22400 mL of $N_{2}$ at STP weighs = 28 gm
Hence the mass of nitrogen which corresponds to 33.53 mL is 22400 mL is $\frac{28 \times 33.53}{224000} = 0.0419 g$

The percentage of nitrogen is $0.0419 \times \frac{100}{0.25} = 16.76 %$

In Duma's method for the 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 Hg, the percentage of nitrogen in the compound is :

Solution:

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