I need assistance on this chem problem

[Equilibrium]

Which has a higher density, oxygen or air? Prove by calculations, knowing that air is approximately 20% oxygen and 80% nitrogen and that 1 mol of any gas occupies 22.4liters at STP

 

[Hootenanny]

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[Blueshift5]

Based on the given information, we can calculate the density of both oxygen and air using the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

For oxygen, we know that it makes up 20% of air, so we can calculate the number of moles of oxygen present in 1 mol of air as: (20/100) x 1 mol = 0.2 mol. Therefore, the number of moles of oxygen in 22.4 liters of air at STP would be: (0.2 mol/1 mol) x 22.4 L = 4.48 L.

Using the ideal gas law, we can calculate the density of oxygen as: (0.2 mol x 0.0821 L atm/mol K x 273.15 K) / 22.4 L = 1.33 g/L.

For air, we know that it is composed of 80% nitrogen, which means that the number of moles of nitrogen present in 1 mol of air is: (80/100) x 1 mol = 0.8 mol. Therefore, the number of moles of nitrogen in 22.4 liters of air at STP would be: (0.8 mol/1 mol) x 22.4 L = 17.92 L.

Using the ideal gas law, we can calculate the density of nitrogen as: (0.8 mol x 0.0821 L atm/mol K x 273.15 K) / 22.4 L = 5.33 g/L.

Since air is a mixture of gases, we can calculate its density by adding the densities of oxygen and nitrogen together: 1.33 g/L + 5.33 g/L = 6.66 g/L.

Therefore, we can conclude that air has a higher density compared to oxygen, with air having a density of 6.66 g/L and oxygen having a density of 1.33 g/L. This is because air is a mixture of gases, including oxygen and nitrogen, and the density of a mixture is the sum of the densities of its individual components.