A sample of 0 . 125 g of an organic compound when analysed by Duma's method yields 22 . 78 mL of nitrogen gas collected over KOH solution at 280 K and 759 mmHg . The percentage of nitrogen in the

Question

A sample of 0 . 125 g of an organic compound when analysed by Duma's method yields 22 . 78 mL of nitrogen gas collected over KOH solution at 280 K and 759 mmHg . The percentage of nitrogen in the given organic compound is____ (a) The vapour pressure of water at 280 K is 14 . 2 mmHg (b) R = 0 . 082 L atm K - 1 mol - 1

TestHub · Accepted Answer

Let the mass of organic compound = m g
Volume of nitrogen collected = V₁mL
Room temperature = T₁K
$V = 22.78 ml, T = 280 K$
$P_{total} = 759 mmHg$
$P_{N_{2}}$ = Atmospheric pressure – Aqueous tension
$P_{N_{2}} = 759 - 14.2 = 744.8 mmHg$
$22400 mL N_{2} at STP weighs 28 g .$
$V mL N_{2} gas at STP weighs = \frac{28 \times V}{22400} g$
$n_{N_{z}} = \frac{744.8 \times 22.78}{760 \times 1000 \times 0.082 \times 280} = 0.00097$
$W_{Nitrogen} = 0.02716$ g
$% N = \frac{0.02716}{0.125} \times 1000 = 21.728$

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