The density of ethane, $C X_{2} H X_{6}$ (molar mass $30 g mo l^{- 1}$ ), at $400 K$ and $1.5 atm$ , taking $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , in $g L^{- 1}$ (three decimals) is

Numerical answer type, answer in g/L, tolerance 0.003.

Answer: 1.37 g/L

d = \frac{PM}{RT} = \frac{1.5 \times 30}{0.0821 \times 400} = \frac{45}{32.84} = 1.37028 \approx 1.370 g L^{- 1}