Question

A mixture of oxygen and argon gases, at a total pressure of 740 mm Hg, contains 6.32 grams of oxygen and 3.81 grams of argon. What is the partial pressure of each gas in the mixture?

P_O2 =  mm Hg
P_Ar =  mm Hg

A mixture of neon and helium gases contains neon at a partial pressure of 181 mm Hg and helium at a partial pressure of 508 mm Hg. What is the mole fraction of each gas in the mixture?

X_Ne =
X_He =

Answer 1

O2, Ar,

PT = 740 mm Hg

m = 6.32 g of Oxygen gas

mol of O2 = mass/MW = 6.32/32 = 0.1975 mol of O2

m = 3.81 g f Argon

MW of Ar = 39.948 g/mol

mol of Ar = mass/MW = 3.81 /39.948 = 0.0953 mol of Ar

molar fractions = mol of O2 / total mol = 0.1975 /(0.0953+0.1975 ) = 0.67452

molar fraction of Ar = 1-x = 1-0.67452 = 0.32548

P-O2 = x-O2*PT = 0.67452*740 = 499.1448 mm Hg

P-Ar = x-Ar*PT = 0.32548 *740 = 240.8552 mm HG

Q2.

P-Ne= 181 mm Hg

P-He = 508 mm Hg

total P = P-He + P-Ne = 181 + 508 = 689 mm Hg

x-Ne = P-He / PT = 181 mm Hg / 689 mm Hg =0.26269

x-He = P-O2/ PT = 508 mm Hg / 689 mm Hg =0.737