Question

In a mixture of He, Ne, and Xe gases with a total pressure of 925 atm, if there is 10.5 g of each gas in the mixture, what is the partial pressure of Xe?

Answer 1

Partial pressure of Xe = 22.9 atm

Step-by-step explanation:

Given:

Total pressure of gas mixture = 925 atm

Mass of He = Ne = Xe = 10.5 g

To determine:

Partial pressure of Xe

Step-by-step explanation:

As per Dalton's Law in a mixture of gases the total pressure is the equal to the sum of partial pressures

In this case:

`Ptotal = p(Xe) + p(He) + p(Ne)\\`-------------(1)

Partial pressure of each gas can be expressed as:

`Partial\ pressure = mole\ fraction * total\ pressure`

`pXe = (n(Xe))/(n(total)) * Ptotal`-----(2)

where n(Xe) = moles of Xe

n(Total) = total moles

`nXe = (mass\ Xe)/(At\ mass\ Xe) = (10.5g)/(131g/mol) =0.080`

`nHe = (mass\ He)/(At\ mass\ He) = (10.5g)/(4g/mol) =2.625`

`nNe = (mass\ Ne)/(At\ mass\ Ne) = (10.5g)/(20g/mol) =0.525`

Therefore,

n(Total) = 0.080+2.625+0.525 = 3.23

Substituting for nXe, n(Total) and P(total) in equation (2)

`pXe = (0.080)/(3.23) * 925 = 22.9 atm`

Answer 2

Answer is: partial pressure of Xe is 22,95 atm.
m(He) = 10,5 g.
n(He) = m(He) ÷ M(He).
n(He) = 10,5 g ÷ 4 g/mol
n(He) = 2,625 mol.
m(Ne) = 10,5 g.
n(Ne) = m(Ne) ÷ M(Ne).
n(Ne) = 10,5 g ÷ 20,18 g/mol.
n(Ne) = 0,52 mol.
m(Xe) = 10,5 g.
n(Xe) = 10,5 g ÷ 131,3 g/mol.
n(Xe) = 0,08 mol.
Using Dalton's law:
p(Xe) = (0,08 mol / 0,08 mol + 0,52 mol + 2,625 mol) · 925 atm.
p(Xe) = 22,95 atm.