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 : [2015]
(a) 18.20 (b) 16.76
(c) 15.76 (d) 17.36

Answer 1

The percentage of nitrogen is 16.76 %.

Option (B) is correct.

Explanation:

Given data:

• Mass "m" = 0.25 g
• V1 = 40 mL
• T1 = 300 K
• P1 = 725 mm - 25 mm = 700 mm
• Po = 760 mm
• To =  273 K  , Vo = ?

Vo = P1 x V1 / T1 x To / Po = 700 mm x 40 ml / 300 K x 273 K / 760 mm

Vo = 33.53 ml

At STP, 22400 ml of nitrogen occupies 22400 ml.

22400 mL OF N 2  at STP weighs = 28 gm

Mass of nitrogen = 28 x  33.53 / 22400 = 0.0419 g

Percentage of Nitrogen = 0.0419 x 100 / 0.25 = 16.76 %

Thus the percentage of nitrogen is 16.76 %.

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