A room of volume $30 m^{3}$ contains air (average molar mass $29 g mo l^{- 1}$ ) at $300 K$ and $1.0 atm$ . Taking $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , the mass of air in kilograms (one decimal) is

Numerical answer type, answer in kg, tolerance 0.1.

Answer: 35.3 kg

V = 30 m^{3} = 3.0 \times 1 0^{4} L

n = \frac{1.0 \times 3.0 \times 1 0 ^{4}}{0.0821 \times 300} = 1218.0 mol

. Mass

= 1218.0 \times 29 = 35323 g = 35.32 \approx 35.3 kg