In Duma's method of estimation of nitrogen, 0.1840 g of an organic compound gave 30 mL of nitrogen collected at 287 K and 758 mm of Hg pressure. The percentage composition of nitrogen in the compound is __________. (Round off to the nearest integer).
[Given: Aqueous tension at 287K = 14 mm of Hg]
In Duma's method of estimation of nitrogen, 0.1840 g of an organic compound gave 30 mL of nitrogen collected at 287 K and 758 mm of Hg pressure. The percentage composition of nitrogen in the compound is __________. (Round off to the nearest integer).
[Given: Aqueous tension at 287K = 14 mm of Hg]
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1 Answer
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The partial pressure of dry N? is 758 - 14 = 744 mm Hg. Using the ideal gas law (PV=nRT), the moles of N? are calculated to be 1.25 × 10? ³ mol. This corresponds to 0.035 g of N? The percentage of nitrogen in the sample is (0.035 g / 0.1840 g) × 100, which is 18.96%.
Answer: 19 (Rounded)
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