Question

gas mixture consists of N2, O2, and Ne, where the mole fraction of N2 is 0.55 and the mole fraction of Ne is 0.25. If the mixture is at STP in a 5.0 L container, how many molecules of O2 are present?

Answer 1

Answer:

2.69x10²² molecules of O₂

Explanation:

We start with the knowldge that says: "the sum of mole fractions in a mixture of gases = 1 ". When we start with this, we propose

Mole fraction N₂ + Mole fraction Ne + Mole fraction O₂ = 1

0.55 + 0.25 + Mole fraction O₂ = 1

Mole fraction O₂ = 1 - 0.25 - 0.55 → 0.20

We know that the mixture is at STP in a 5L of volume, so let's calculate the volume. We use the Ideal Gases Law

1 atm . 5L = n . 0.082 . 273 K

n = 1 atm . 5L / 0.082 . 273K → 0.223 moles

These number of moles are the total moles in the mixture. We apply the mole fraction to determine the moles of O₂ that are present in the container.

Mole fraction O₂ = Moles O₂ / Total moles → Mole fraction O₂ . Total moles = moles O₂

0.223 moles . 0.2 = 0.0446 moles of O₂

Let's count the number of molecules

0.0446 moles . 6.02x10²³molecules / 1 mol = 2.69x10²² molecules of O₂

Answer 2

Answer:

We have 2.69*10^22 O2 molecules

Explanation:

Step 1: Data given

Mol fraction of N2 = 0.55

Mol fraction of Ne = 0.25

Mol fraction O2 = 1- 0.55 - 0.25 = 0.20

Volume = 5.0 L

Step 2: Calculate total number of moles

p*V = n*R*T

⇒with P = the pressure at STP = 1 atm

⇒with V = the volume = 5.0 L

⇒with n = the number of moles = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 K*atm/mol * K

⇒with T = the temperature = 273 K

n = (p*V) / (R*T)

n = (1 * 5) / (0.08206*273)

n = 0.2232 moles

Step 3: Calculate moles O2

0.2232 moles * 0.2 = 0.04464 moles

Step 4: Calculate molecules O2

0.04464 moles * 6.022 *10^23 / moles

2.69*10^22 molecules

We have 2.69*10^22 O2 molecules