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)

Step by Step Solution

Step 1: Pressure of dry nitrogen gas is obtained by subtracting aqueous tension.
Dry pressure of N₂ = 715 − 15 = 700 mm Hg

Step 2: Convert pressure into atmosphere.
700 mm Hg = 700 / 760 atm = 0.921 atm

Step 3: Convert volume into litres.
60 mL = 0.060 L

Step 4: Use ideal gas equation:
PV = nRT

Step 5: Substitute values:
n = (P × V) / (R × T)
n = (0.921 × 0.060) / (0.0821 × 300)

Step 6: n = 0.05526 / 24.63
n ≈ 0.00224 moles of N₂

Step 7: Moles of nitrogen atoms = 2 × 0.00224 = 0.00448 moles

Step 8: Mass of nitrogen = 0.00448 × 14 = 0.0627 g

Step 9: Percentage of nitrogen =
(0.0627 / 0.4) × 100

Step 10: Percentage of nitrogen ≈ 15.7%

✅ Final Answer: 15.71%