In Dumas' method for estimation of nitrogen 0.4 g of an organic compound gave 60 mL of nitrogen collected at 300 K temperature and 715 mm Hg pressure. The percentage composition of nitrogen in the compound is (Given : Aqueous tension at 300 K=15 mm Hg )

Solution

Chemistry ma’am · 4 steps

Dry the gas first, take it to STP, then let 22400 mL weigh 28 g of nitrogen.

1. Remove water vapour. The N₂ is collected over water, so the measured pressure includes water vapour. The dry gas pressure is 715 - 15 = 700 mm Hg.
2. Convert to STP. Use P₁V₁/T₁ = P₂V₂/T₂ with STP as 760 mm Hg and 273 K.

   V_(STP) = 60 × 700/760 × 273/300 = 50.29 mL

3. Mass of nitrogen. At STP, 22400 mL of N₂ weighs 28 g. So the mass is (28 × 50.29)/22400 = 0.06286 g.
4. So. Percentage of N = 0.06286/0.4 × 100 = 15.71%, which is option (a).

The idea to keep. In Dumas' method, subtract the aqueous tension before anything else, then take the gas to STP. The gas is N₂, so 22400 mL weighs 28 g, not 14 g.

Answer from the NTA final answer key. Solution by PrepWiser.