Question

Assuming that only the listed gases are present, what would be the mole fraction of oxygen gas be for each of the following situations?

A gas sample of 2.31 atm of oxygen gas and 3.75 atm of hydrogen gas that react to form water vapor. Assume the volume of the container and the temperature inside the container does not change.​

Answer 1

The mole fraction of oxygen gas : 0.381

Further explanation

Given

2.31 atm Oxygen

3.75 atm Hydrogen

Required

The mole fraction of Oxygen

Solution

Dalton Law's of partial pressure

P tot = P₁ + P₂ + ..Pₙ

Input the value :

P tot = P O₂ + P H₂

P tot = 2.31 atm + 3.75 atm

P tot = 6.06 atm

The mole fraction of O₂ (X O₂):

P O₂ = X O₂ x P tot

X O₂ = P O₂ / P tot

X O₂ = 2.31 /6.06

X O₂ = 0.381

Answer 2

The mole fraction of oxygen gas in the given mixture of oxygen and hydrogen gas is 0.381.

How do we calculate mole fraction?

If the details is present in terms of partial pressure then by using the below equation, we can calculate the mole fraction as:

p = (x)(P), where

p = partial pressure

P = total pressure

x = mole fraction

Total pressure of the mixture will be calculated by adding the partial pressure of oxygen and hydrogen gases as:

P = 2.31 + 3.75 = 6.06 atm

Now mole fraction of oxygen will be calculated as:

x = 2.31 / 6.06 = 0.381

Hence mole fraction of oxygen is 0.381.

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