A cylinder of volume $2.0 L$ contains nitrogen at $100 atm$ and $300 K$ , where $Z = 1.25$ . The mass of nitrogen in the cylinder, in grams (one decimal), is (molar mass of $N X_{2}$ $= 28 g mo l^{- 1}$ , $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ )

Numerical answer type, answer in g, tolerance 0.2.

Answer: 181.9 g

n = \frac{P V}{ZRT} = \frac{100 \times 2.0}{1.25 \times 0.0821 \times 300} = \frac{200}{30.7875} = 6.4961 mol

. Mass

= 6.4961 \times 28 = 181.89 g

, i.e.

181.9 g

. Treating the gas as ideal would overestimate the mass by 25 percent.