What is the mole fraction of oxygen in a gas mixture that is 22% oxygen and 78% nitrogen by volume?
A)0)20
B)0)22
C)0)25
D)0)28

Question

Quizplus · Accepted Answer

The mole fraction of a gas is the ratio of the number of moles of that gas to the total number of moles of all gases in the mixture. In this case, the mole fraction of oxygen is:$$X_{O_2} = \frac{P_{O_2}}{P_{total}}$$where $P_{O_2}$ is the partial pressure of oxygen and $P_{total}$ is the total pressure of the gas mixture.The partial pressure of oxygen is:$$P_{O_2} = X_{O_2} 	imes P_{total}$$where $X_{O_2}$ is the mole fraction of oxygen and $P_{total}$ is the total pressure of the gas mixture.The total pressure of the gas mixture is:$$P_{total} = P_{O_2} + P_{N_2}$$where $P_{N_2}$ is the partial pressure of nitrogen.The partial pressure of nitrogen is:$$P_{N_2} = X_{N_2} 	imes P_{total}$$where $X_{N_2}$ is the mole fraction of nitrogen and $P_{total}$ is the total pressure of the gas mixture.Substituting the expressions for $P_{O_2}$ and $P_{N_2}$ into the expression for $P_{total}$, we get:$$P_{total} = X_{O_2} 	imes P_{total} + X_{N_2} 	imes P_{total}$$Simplifying, we get:$$P_{total} = (X_{O_2} + X_{N_2}) 	imes P_{total}$$Dividing both sides by $P_{total}$, we get:$$1 = X_{O_2} + X_{N_2}$$Rearranging, we get:$$X_{O_2} = 1 - X_{N_2}$$Substituting the given values for $X_{O_2}$ and $X_{N_2}$, we get:$$X_{O_2} = 1 - 0.78 = 0.22$$Therefore, the mole fraction of oxygen in the gas mixture is 0.22.

What Is the Mole Fraction of Oxygen in a Gas