# Question 90bc3

Jun 9, 2017

Given

Volume of nitrogen $V = 82.1 m L = 0.0821 L$

Pressure of the gas collected excluding aqueous tension

$P = \left(774.5 - 14.5\right) m m = \frac{760}{760} = 1 a t m$

Temperature of the gas $T = \left(27 + 273\right) = 300 K$

Universal gas constant $R = 0.0821 L a t m {K}^{-} 1 m o {l}^{-} 1$

Let mass of the gas be $w$ g

Molar mass of the gas $M = 2 \times 14 = 28 g \text{/} m o l$

So applying Ideal gas equation

$P V = \frac{w}{M} R T$

$\implies w = \frac{P V M}{R T}$

$\implies w = \frac{1 \times 0.0821 \times 28}{0.0821 \times 300} = \frac{28}{300} g$

So percentage of nitrogen in the compound

=(28/300)/0.14xx100=66.7%#