Exercises
Exercise 1. A gas mixture has the following composition:
25.0% H2
2.8% HF
5.4% HCl
1.7% SO2
0.1% H2S.
What is the partial pressure of each of these gases if the total pressure of the gas mixture were 759 mmHg?
The percentage of each gas is given. These are the mole percents. If we divide the percentages by 100, the fractions add to one. These are the mole fractions.
CO2: 0.650 x 759 mmHg = 493 mmHg
H2: 0.250 x 759 mmHg = 190. mmHg
HF: 0.028 x 759 mmHg = 21 mmHg
HCl: 0.054 x 759 mmHg = 41 mmHg
SO2: 0.0178 x 759 mmHg = 14 mmHg
H2S: 0.001 x 759 mmHg = 0.76 mmHg
Back to Dalton’s Law of Partial Pressures
Exercise 2. Both 1.23 mg of O2 and 0.51 mg of He are contained in a 250.00 mL flask at 18.0°C. Calculate the partial pressures of both gases. Calculate the total pressure in the flask.
First, we convert mg to grams and then to moles. 1.23 mg of O2 = 3.84 x 10-5 moles and 0.51 mg of He = 1.27 x 10-4 moles.
Determine the mole fraction of O2
\(\displaystyle \chi_{O_2}\;=\;\frac{3.84\times\;10^{-5}\;mol}{3.84\times\;10^{-5}\;mol\;+\;1.27\times\;10^{-4}\;mol}\;=\;0.232\)
First, find the total pressure using PV = nRT — solve for P
V = 0.25000 L
n = 3.84 x 10-5 mol + 1.27 x 10-4 mol = 1.65 x 10-4 moles,
T = 18.0 + 273.15 = 291.15 K
R = \(0.0821\;\frac{L⋅atm}{mol⋅K}\)
\(\displaystyle P_{T}\;=\;\frac{nRT}{V}\;=\;\frac{1.65\;\times10^{-4}\;mol\times\;0.0821\frac{L⋅atm}{mol⋅K}\times\;291.15\;K}{0.25000\;L}\;=\;0.0158\;atm\)
Next, we find the partial pressure of O2.
PO2 = χO2PT = 0.232 x 0.0158 atm = 0.00367 atm
Next, we subtract the mole fraction of O2 from 1. The mole fractions sum to 1.
1 – 0.232 = 0.768
PHe = 0.768 x 0.0158 = 0.0121 atm
Back to Dalton’s Law of Partial Pressures
Exercise 3. A mixture of Ar and N2 gasses have a density of 1.435 g/L at STP. Calculate the mole fraction of each gas.
At STP one mole of gas occupies a volume of 22.4 L. The mass of one mole Ar is 39.948 g and the mass of 1 mole N2 is 28.0134 g.
The mass of 22.4 L of the mixture is:
\(\displaystyle 22.4\;L\;\times 1.435\frac{g}{L}\;=\;32.14\;g\)
The mass of the mixture is the weighted average of molar masses and we can let x be the mole fraction of Ar and (1 – x) the mole fraction of N2.
(x)(39.948 g/mol) + (1 – x)(28.0134 g/mol) = 32.14 g
39.948x + 28.0134 – 28.0134x = 32.14 g
Solving for x, we have 11.93x = 4.127 and x = 0.346
The mole fraction for Ar is 0.346
The mole fraction for N2 is 1 – 0.346 = 0.654
Back to Dalton’s Law of Partial Pressures
Exercise 4. Chlorine, Cl2, gas can be formed by the following reaction:
2 NaCl (s) + 2 H2SO4 (l) + MnO2 (s) → Na2SO4 (s) + MnSO4 (s) + 2 H2O (g) + Cl2 (g)
A volume of 0.602 L of gas is collected over water at 28.0°C and a pressure of 758 mmHg. Calculate the mole fraction of Cl2.
PT = PCl2 + PH2O = 758 mmHg
The vapor pressure of water at 28.0°C is 28.7 mmHg.
PCl2 = PT – PH2O = 755 mmHg – 28.7 mmHg = 726 mmHg
\(\displaystyle \chi_{Cl_2} = \frac{P_{Cl_2}}{P_T}\;=\;\frac{726\;mmHg}{755\;mmHg}\;=\;\mathbf{0.962\;mmHg}\)