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?

The number of molecules of O₂ present in the container which consists of N₂, O₂ and Ne is 2.68×10²² molecules

Mole fraction of N₂ = 0.55

Mole fraction of Ne = 0.25

Total mole fraction = 1

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

0.55 + 0.25 + Mole fraction of O₂ = 1

0.8 + Mole fraction of O₂ = 1

Collect like terms

Mole fraction of O₂ = 1 – 0.8

Pressure (P) = 1 atm (at stp)

Temperature (T) = 273 (at stp)

Volume (V) = 5 L

Gas constant (R) = 0.0821 atm.L /Kmol

1 × 5 = n × 0.0821 × 273

5 = n × 22.4133

Divide both side by 22.4133

Thus, the total mole  of the gases in the container is 0.223 mole

Mole fraction of O₂ = 0.2

Total mole = 0.223 mole

Mole = Total mole × mole fraction

Mole of O₂ = 0.2 × 0.223

From Avogadro's hypothesis

1 mole of O₂ = 6.02×10²³ molecules

Therefore,

0.0446 mole of O₂ = 0.0446 × 6.02×10²³

0.0446 mole of O₂ = 2.68×10²² molecules

Thus, the number of molecules of O₂ present in the container is 2.68×10²² molecules

Learn more: link

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

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm