In a Dumas estimation, 0.24 g of a compound gave 36.0 mL of moist nitrogen collected over water at 27 $^{\circ}$ C and a total pressure of 756 mm Hg. Aqueous tension at 27 $^{\circ}$ C is 26 mm Hg. Take 273 K, 760 mm Hg, 22400 mL per mole at STP and N = 14. The percentage of nitrogen is

Answer: B

Pressure of dry

N X_{2}

= 756 - 26 = 730 mm Hg. Reducing to STP:

V_{0} = \frac{730 \times 36.0 \times 273}{300 \times 760} = 31.47

mL. Mass of N =

\frac{28}{22400} \times 31.47 = 0.03933

g, so %N =

\frac{0.03933}{0.24} \times 100 = 16.39%