$ 0.2325 \mathrm{~g} $ of an organic compound was analysed for nitrogen by Dumas method. $ 31.7 \mathrm{~mL} $ of moist nitrogen was collected at $ 25^{\circ} \mathrm{C} $ arid $ 755.8 \mathrm{~mm} $ of $ \mathrm{Hg} $ pressure. Calculate the percentage of nitrogen in the sample. (Aqueous tension of water at $ 25^{\circ} \mathrm{C} $ is $ 23.8 \mathrm{~mm} $ of $ \mathrm{kg} $ )

Question

$ 0.2325 \mathrm{~g} $ of an organic compound was analysed for nitrogen by Dumas method. $ 31.7 \mathrm{~mL} $ of moist nitrogen was collected at $ 25^{\circ} \mathrm{C} $ arid $ 755.8 \mathrm{~mm} $ of $ \mathrm{Hg} $ pressure. Calculate the percentage of nitrogen in the sample. (Aqueous tension of water at $ 25^{\circ} \mathrm{C} $ is $ 23.8 \mathrm{~mm} $ of $ \mathrm{kg} $ ), $ 18 \% $, $ 19.7 \% $, $ 15.1 \% $, $ 20.2 \% $

MARKS App · Accepted Answer

At given condition, P 1 = 755 . 8 - 23 . 8 = 732 . 0 mm V 1 = 31 . 7 mL T 1 = 25 + 273 = 298 K For STP , P 2 = 760 mm T 2 = 273 K R 2 = ? Applying the gas equation, P 1 × V 1 T 1 = P 2 × V 2 T 2 732 × 31 . 7 298 = 760 × V 2 273 V 2 = 28 mL % of N = 28 × 28 × 100 22400 × 0 . 2375 = 15 . 1 %

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