Question

A gas mixture contains the following gases at the indicated partial pressures: N2, 310. torr ; O2, 144 torr ; and He, 225 torr. 1. What mass of each gas is present in a 2.85 L sample of this mixture at 25.0 ∘C?

Answer 1

The total pressure is 310 + 144 + 225 = 679 Torr.

The fraction represented by the N2 is 310/679 = 0.456; the fraction for O2 is 144/679 = 0.212; the fraction for He is 225/679 = 0.331.

Convert the total pressure to atmospheres:
(679 Torr)(0.00132 atm/Torr) = 0.896 atm

Convert the fractional pressures to atmospheres:
(0.456)(0.896) = 0.408 atm N2.
(0.212)(0.896) = 0.189 atm O2.
(0.331)(0.896) = 0.296 atm He.

Now use PV = nRT. This gives n = PV/(RT).

V = 2.85 L total.
V (N2) = (0.456)(2.85 L) = 1.299 L.
V (O2) = (0.212)(2.85 L) = 0.604 L.
V (He) = (0.331)(2.85 L) = 0.943 L.

T= 25 °C = 298 K.(Temperature)

R = 0.082 L·atm/(mol·K).

n(N2) = P(N2)·V(N2)/(RT) [PV=nRT]
= (0.408 atm)(1.299 L) / ((0.082 L·atm/(mol·K))(298 K))
= 0.02168 mol.

n(O2) = P(O2)·V(O2)/(RT)
= (0.189 atm)(0.604 L) / ((0.082 L·atm/(mol·K))(298 K))
= 0.00467 mol.

n(He) = P(He)·V(He)/(RT)
= (0.296 atm)(0.943 L) / ((0.082 L·atm/(mol·K))(298 K))
= 0.01142 mol.

The molar mass of N2 is 28.0 g/mol; that of O2 is 32.0 g/mol; that of He is 4.00 g/mol. Multiply these by the computed number of moles:
N2: (0.02168 mol)(28.0 g/mol) = 0.607 g.
O2: (0.00467 mol)(32.0 g/mol) = 0.1490 g.
He: (0.01142 mol)(4.00 g/mol) = 0.0456 g.

Mass of N2=0.607 g, O2=0.149 g and He=0.0456 g.