The density of butane, $C X_{4} H X_{10}$ (molar mass $58 g mo l^{- 1}$ ), at $350 K$ and $2.0 atm$ , taking $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , in $g L^{- 1}$ (two decimals) is

Numerical answer type, answer in g/L, tolerance 0.02.

Answer: 4.04 g/L

d = \frac{PM}{RT} = \frac{2.0 \times 58}{0.0821 \times 350} = \frac{116}{28.735} = 4.0369 \approx 4.04 g L^{- 1}