What will be the pressure of a sample of 48.0 grams of oxygen gas in a glass container of volume 5.2 L at 25 and degC?

To find the pressure of the oxygen gas, we can use the Ideal Gas Law: ( PV = nRT ). First, we need to calculate the number of moles of oxygen gas (O₂): ( n = \frac{48.0 , \text{g}}{32.00 , \text{g/mol}} = 1.5 , \text{mol} ). At 25°C (or 298 K), using the ideal gas constant ( R = 0.0821 , \text{L·atm/(K·mol)} ), we can rearrange the equation to solve for pressure ( P ):

[ P = \frac{nRT}{V} = \frac{(1.5 , \text{mol})(0.0821 , \text{L·atm/(K·mol)})(298 , \text{K})}{5.2 , \text{L}} \approx 7.13 , \text{atm}. ]

Thus, the pressure of the oxygen gas in the container is approximately 7.13 atm.