$8.0 g$ of $O X_{2}$ (molar mass $32 g mo l^{- 1}$ ) is placed in a $4.1 L$ vessel at $300 K$ . Taking $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , the pressure of the gas in atm (two decimals) is

Numerical answer type, answer in atm, tolerance 0.01.

Answer: 1.5 atm

n = 8.0/32 = 0.25 mol

P = \frac{n RT}{V} = \frac{0.25 \times 0.0821 \times 300}{4.1} = 1.5018 \approx 1.50 atm