Question

a gaseous mixture composed of 16.0 g of o_{2}, 21.0 g of n_{2}, and 16.0 g of he at 20.0°c is confined in a 25.0 l container. what is the partial pressure of o_{2} in this mixture (in atm)?

Answer 1

To find the partial pressure of O₂ in a gaseous mixture, we need to calculate the mole fraction of O₂ and then use Dalton's law of partial pressures.

Step-by-step explanation:

To find the partial pressure of O₂ in the given mixture, we need to calculate the mole fraction of O₂ and then use Dalton's law of partial pressures.

First, calculate the moles of each gas in the mixture:

• O₂: mass of O₂ / molar mass of O₂ = 16.0 g / 32.0 g/mol = 0.5 mol
• N₂: mass of N₂ / molar mass of N₂ = 21.0 g / 28.0 g/mol = 0.75 mol
• He: mass of He / molar mass of He = 16.0 g / 4.0 g/mol = 4.0 mol

Next, calculate the total moles of gas in the mixture:

Total moles = moles of O₂ + moles of N₂ + moles of He = 0.5 mol + 0.75 mol + 4.0 mol = 5.25 mol

Now, calculate the mole fraction of O₂:

Mole fraction of O₂ = moles of O₂ / total moles = 0.5 mol / 5.25 mol = 0.0952

Finally, use Dalton's law of partial pressures:

Partial pressure of O₂ = mole fraction of O₂ * total pressure = 0.0952 * total pressure

Since the total pressure is not given, we cannot calculate the exact partial pressure of O₂. Please provide the total pressure in order to calculate the partial pressure of O₂ in this mixture.

Answer 2

The partial pressure of the oxygen is 0.48 atm.

Partial pressure is a concept in thermodynamics that relates to the pressure exerted by a particular gas in a mixture of gases.

Number of moles of oxygen = 16.0 g/32 g/mol = 0.5 moles

Number of moles of nitrogen = 16.0 g /28 g/mol = 0.57 moles

Number of moles of He = 16 g/4 g/mol = 4 moles

Total number of moles = 0.5 moles + 0.57 moles + 4 moles

= 5.07 moles

To find the total pressure;

PV = nRT

P = nRT/V

P = 5.07 * 0.082 * 293/25

P = 4.87 atm

Partial pressure = Mole fraction * Total pressure

= 0.5 moles /5.07 moles * 4.87 atm

= 0.48 atm