In Dumas'method of estimation of nitrogen 0.35 g of an organic compound gave 55 mL of nitrogen collected at 300 K temperature and 715 mm pressure. The percentage composition of nitrogen in the compound would be:
(Aqueous tension at 300 K = 15 mm)

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14.45

15.45

16.45

17.45

answer is C.

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Detailed Solution

$P_{N_{2}} = 715 - 15 = 700 mm V = 55 mL ∴ PV = \frac{w}{M} RT \frac{700}{760} \times \frac{55}{1000} = \frac{w_{N_{2}}}{28} \times 0.0821 \times 300 w_{N_{2}} = 0.058 g ∴ % N_{2} = \frac{0.058}{0.35} \times 100 = 16.45$

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