A $5.0 L$ cylinder contains $N X_{2}$ (molar mass 28) at $300 K$ and $2.0 atm$ . Taking $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , the mass of $N X_{2}$ in grams (one decimal) is

Numerical answer type, answer in g, tolerance 0.1.

Answer: 11.4 g

n = \frac{P V}{RT} = \frac{2.0 \times 5.0}{0.0821 \times 300} = 0.4060 mol

. Mass

= 0.4060 \times 28 = 11.368 \approx 11.4 g