A mixture of #"N"_2#, #"O"_2#, and #"Ar"# has mole fractions of #"N"_2 = .780#, #"O"_2 = .210#, and #"Ar" = .010#. What is the partial pressure of #"O"_2# if the total pressure of the mixture is #"750 torr"#?

If the total number of moles in the mixture is #1.50#, how many moles of #"O"_2# are there?

1 Answer
Jul 27, 2017

Answer:

Here's what I got.

Explanation:

The idea here is that the partial pressures of the three gases will depend on the mole fraction each gas has in the mixture #-># think Dalton's Law of Partial Pressures here.

More specifically, you can say that the partial pressure of a gas #i# that's part of a gaseous mixture that has a total pressure of #P_'total"# will be equal to

#P_i = chi_i * P_"total"#

Here #chi_i# represents the mole fraction of gas #i# in the mixture.

In your case, you know that

#chi_ ("O"_ 2) = 0.210#

and that the total pressure of the gas is equal to #"750 torr"#, so you can say that the partial pressure of oxygen gas in this mixture will be equal to

#P_ ("O"_ 2) = 0.210 * "750 torr"#

#P_ ("O"_ 2) = color(darkgreen)(ul(color(black)("160 torr")))#

The answer must be rounded to two sig figs, the number of sig figs you have for the total pressure of the mixture

To find the number of moles of oxygen gas present in the mixture, use the fact that the mole fraction of oxygen is defined as

#chi_ ("O"_ 2) = "no. of moles of O"_2/"total number of moles of gas"#

In your case, you have

#0.210 = "no. of moles of O"_2/"1.50 moles of gas"#

which gets you

#"no. of moles of O"_2 = 0.210 * "1.50 moles"#

#"no. of moles of O"_2 = color(darkgreen)(ul(color(black)("0.315 moles O"_2)))#

This time, the answer is rounded to three sig figs because you're not using the total pressure of the mixture in this calculation.