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?
we know that
moles = mass / molar mass
moles of He = 10.5 g / ( 4 g/mol) = 2.625 mol
moles of Ne = 10.5 g / ( 20 g/mol) = 0.525 mol
moles of Xe = 10.5 g / (131.293 g/mol) = 0.08 mol
total moles of gases = 2.625 + 0.525 + 0.08 = 3.23 mol
mole fraction of Xe = moles of Xe / total moles of gases
mole fraction of Xe = 0.08 mol / 3.23 mol
mole fraction of Xe = 0.024768 mol
partial pressure of Xe = mole fraction of Xe x total pressure
partial pressure of Xe = 0.024768 x 925 atm
partial pressure of Xe = 22.91 atm
Get Answers For Free
Most questions answered within 1 hours.