A sample of 0.125g of an organic compound when analyzed by Duma's method yields 22.78 mL of nitrogen gas collected over KOH solution at 280 K and 759 mm Hg. The percentage of nitrogen in the given organic compound is__________. (Nearest integer)

[Given: (a) The vapour pressure of water of 280K is 14.2mm Hg

(b) R = 0.082 L atm K^-1 mol^-1]

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Vishal Baghel | Contributor-Level 10

6 months ago

$V_{N_{2}} = 22.78 m l T = 280 K$

PTotal = 759 mm of Hg

$P_{N_{2}} = 759 - 14.2 = 744.8 m m H g$

$n_{N_{2}} = \frac{744.8 * (22.78 / 1000)}{(0.082 * 760) * 280}$

= 0.00097

$% N = \frac{0.02716}{0.125} * 100 = 21.728$

$\approx 22 .$

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