A sample of 0.2 g of an organic compound when analyzed by Duma’s method yields 22.4 mL of N₂ gas which is collected over KOH solution at $17^{0} C$ and at a pressure of 749 mm Hg. The percentage of nitrogen in the given organic compound is_____(Nearest integer)
a) Aqueous tension of water at 290 k is 13 mm Hg
b) R=0.0821 L atm/ k.mole.

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answer is 13.

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Detailed Solution

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} \frac{760 \times V_{1}}{273} = \frac{(749 - 13) (22.4)}{290} \frac{760 \times V_{1}}{273} = \frac{736 \times (22.4)}{290} V_{1} = \frac{736}{760} \times 22.4 \times \frac{273}{290} V_{1} = 0.96842 \times 0.9413 \times 22.4 = 0.911 \times 22.4$

$% N = \frac{28 \times 20.42}{22, 400 \times w t . o f o r g a n i c s a m p l e} = \frac{28 \times 20.42}{224 \times 0.2} = \frac{28 \times 20.42}{224 \times 0.2} = \frac{14 \times 20.42}{224} = 12.76 = 13$