Question
In Dumas method for estimation of nitrogen, 0.50 g of an organic compound gave 70 mL of nitrogen collected at 300 K and 715 mm pressure. The percentage of nitrogen in the organic compound is \_\_\_\_ \%. (Aqueous tension at 300 K is 15 mm).
JEE Main
Chemistry
General Organic Chemistry
2026
JEE Main 2026 (24 January Shift 1)
JEE Main Chemistry Question (2026) — Solution
Options
- A. A
- B. B
- C. C
- D. D
Answer
A. A
Step-by-step solution
Given data: Mass of organic compound (w) = 0.50 g Volume of nitrogen gas (V_1) = 70 mL Temperature (T_1) = 300 K Total pressure = 715 mm Hg Aqueous tension = 15 mm Hg Pressure of dry nitrogen (P_1) = 715 - 15 = 700 mm Hg Convert the volume of nitrogen to STP conditions (P_0 = 760 mm Hg, T_0 = 273 K): Using the combined gas law: P_1 V_1 T_1 = P_0 V_0 T_0 V_0 = P_1 V_1 T_0 P_0 T_1 = 700 70 273 760 300 58.68 mL At STP, 22400 mL of N_2 weighs 28 g. Mass of N_2 collected = 28 58.68 22400 g Percentage of nitrogen = Mass of N_2 Mass of compound 100 \% N = 28 58.68 22400 0.50 100 \% N = 28 58.68 112 = 58.68 4 = 14.67 \% Rounding to the nearest integer, we get 15.
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