A gas has density $12.60 g L^{- 1}$ at $10.0 atm$ and $300 K$ , where its compressibility factor is $0.95$ . Using $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , its molar mass in $g mo l^{- 1}$ (one decimal) is

Numerical answer type, answer in g/mol, tolerance 0.2.

Answer: 29.5 g/mol

Z = \frac{PM}{ρRT}

, so

M = \frac{ZρRT}{P} = \frac{0.95 \times 12.60 \times 0.0821 \times 300}{10.0} = \frac{294.821}{10.0} = 29.482

, i.e.

29.5 g mo l^{- 1}

. Ignoring

Z

would give

31.03 g mo l^{- 1}

, an overestimate.