In Duma’s method of estimation of nitrogen, 0.35g of an organic compound gave 55ml of $N_{2}$ collected at 300K and 715 mm pressure. The percentage composition of $N_{2}$ in the compound would be (Aqueous tension at 300K = 15mm)____

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answer is 16.

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Detailed Solution

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}, (\frac{715 - 15}{760}) \frac{55}{300} = \frac{1 \times V}{273}$
$V = 46.09 m l$
$∴ % N = \frac{V}{8 \times w t o f 0 r g . c o m p}$
$= \frac{46.09}{8 \times 0.35} = 16.45 %$

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