In Dumas' method for estimation of nitrogen, 0.3 g of an organic compound gave 50 mL. of nitrogen collected at 300 K temperature and 715 mm pressure. What will be the percentage composition of nitrogen in the compound? (Aqueous tension a 300 K= 15 mm)

50 mL of nitrogen are collected at 300 K and 715 mm of pressure; the actual pressure is 715-15, which equals 700 mm.

the amount of nitrogen at $\mathrm{STP}=\frac{273\times 700\times 50}{300\times 760}$

22,400 mL of ${\mathrm{N}}_2$ at STP weighs = 28g

Nitrogen weighs 41.9 mL. $=\frac{28\times 41.9}{22400}\mathrm{g}$

proportion of nitrogen $=\frac{28\times 41.9\times 100}{22400\times 0.3}=17.46\%$