A gas mixture with a total pressure of 745 mmHg contains each of the following gases at the indicated partial pressures: CO2, 257 mmHg; Ar, 123 mmHg; and O2, 160 mmHg. The mixture also contains helium gas.

What is the partial pressure of the helium gas?

What mass of helium gas is present in a 10.1-L sample of this mixture at 263 K?

To find the partial pressure of the helium gas in the gas mixture, we need to use Dalton's Law of Partial Pressures, which states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of each individual gas.

In this case, the total pressure of the gas mixture is given as 745 mmHg, and the partial pressures of CO2, Ar, and O2 are given as 257 mmHg, 123 mmHg, and 160 mmHg respectively.

To find the partial pressure of helium, we can subtract the sum of the partial pressures of CO2, Ar, and O2 from the total pressure of the mixture:

Partial pressure of helium = Total pressure of mixture - (Partial pressure of CO2 + Partial pressure of Ar + Partial pressure of O2)
Partial pressure of helium = 745 mmHg - (257 mmHg + 123 mmHg + 160 mmHg)
Partial pressure of helium = 745 mmHg - 540 mmHg
Partial pressure of helium = 205 mmHg

Therefore, the partial pressure of helium gas in the mixture is 205 mmHg.

To find the mass of helium gas present in a 10.1-L sample of this mixture at 263 K, we can use the Ideal Gas Law, which states that the number of moles of a gas is directly proportional to its volume, temperature, and pressure. The equation for the Ideal Gas Law is:

PV = nRT

Where:
P = Pressure of the gas (in atm)
V = Volume of the gas (in liters)
n = Number of moles of the gas
R = Ideal Gas Constant (0.0821 L.atm/mol.K)
T = Temperature of the gas (in Kelvin)

First, we need to convert the given pressure from mmHg to atm:

Partial pressure of helium = 205 mmHg * (1 atm / 760 mmHg) = 0.270 atm

We also need to convert the given temperature from degrees Celsius to Kelvin:

Temperature in Kelvin = 263 K

Now we have the pressure (0.270 atm), volume (10.1 L), and temperature (263 K). We can rearrange the Ideal Gas Law equation to solve for the number of moles:

n = PV / RT

n = (0.270 atm * 10.1 L) / (0.0821 L.atm/mol.K * 263 K)
n = 2.9998 moles

Finally, we can find the mass of helium gas by multiplying the number of moles by the molar mass of helium, which is approximately 4.003 grams/mole:

Mass of helium gas = 2.9998 moles * 4.003 grams/mole
Mass of helium gas ≈ 11.999 grams

Therefore, the mass of helium gas present in a 10.1-L sample of this gas mixture at 263 K is approximately 11.999 grams.