## General Chemistry: Principles and Modern Applications (10th Edition)

Molarity = $2.34 \times 10^{-4} \space M \space O_2$
1. Find the amount of moles of $O_2$: $$Volume = 5.77 \space mL \times \frac{1 \space L}{1000 \space mL} = 5.77 \times 10^{-3} \space L$$ $$Temperature/K = 25 + 273.15 = 300 \space K$$ - According to the Ideal Gas Law: $$n = \frac{PV}{RT} = \frac{( 1.00 \space atm )( 5.77 \times 10^{-3} \space L )}{( 0.08206 \space atm \space L \space mol^{-1} \space K^{-1} )( 300 \space K)}$$ $$n = 2.34 \times 10^{-4} \space mol$$ Therefore: Molarity = $2.34 \times 10^{-4} \space mol/L$