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, 18.20, 16.76, 14.76, 20.36

MARKS App · Accepted Answer

Volume of nitrogen collected at 300 K and 725 mm pressure is 40 mL actual pressure = 725 - 25 = 700 m m Using P 1 V 1 G i v e n c o n d i t i o n s T 1 = P 2 V 2 A t S T P T 2 V 2 = P 1 V 1 × T 2 T 1 × P 2 Volume of nitrogen at STP ( V 2 ) = 273 × 700 × 40 300 × 760 = 33.52 m L As 22,400 mL of N 2 at STP weight = 28 g So, 33.5 mL of nitrogen weight = 28 × 33.52 22400 g N% = 28 × V 2 × 100 22400 × W Percentage of nitrogen (N%) = 28 × 33.52 × 100 22400 × 0.25 = 16.76 %

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