A gas has density $0.880 g L^{- 1}$ at $350 K$ and $1.00 atm$ . Taking $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , its molar mass in $g mo l^{- 1}$ (one decimal) is

Numerical answer type, answer in g/mol, tolerance 0.1.

Answer: 25.3 g/mol

M = \frac{d RT}{P} = 0.880 \times 0.0821 \times 350 = 25.287 \approx 25.3 g mo l^{- 1}

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