Question

Air contains ${{N}_{2}}$ and ${{O}_{2}}$ in $4 : 1$ ratio by volume. Approximately molecular weight of air (g/mole):
A.) 29
B.) 38
C.) 16
D.) 74

Answer 1

Hint: Air is containing both the gases that are Nitrogen and Oxygen is a certain ratio. The formula to calculate the average molecular weight of a mixture is summation of the product of the mole fraction and the molecular weight of the respective compound.

Complete Solution :
The air is having a mixture of nitrogen and oxygen gases. Therefore the molecular weights of the air will also change according to the ratio of the nitrogen and the oxygen present in the air.
Given that the ratio of the nitrogen and oxygen present in air is 4:5 that means nitrogen presents 4 parts out of 5 and oxygen contains 1 part out of 5 parts.

Therefore , The mole fraction of the nitrogen is $\dfrac{4}{5} = 0.8$
The mole fraction of the oxygen is $\dfrac{1}{5} = 0.2$
We know that the molecular weight of the nitrogen is = $28 amu$
The molecular weight of the nitrogen is = $32 amu$

The average molecular weight of the air is = (mole fraction of the nitrogen) (molecular weight of the nitrogen) + (mole fraction of the oxygen) (molecular weight of the oxygen)
Put the above values in the above equation and calculate the average molecular weight of the air
Weight = $0.8\times 28 + 0.2\times 32 = 29g/mol$
Therefore the required molecular weight of the air is = $29g/mol$
So, the correct answer is “Option A”.

Note: Not only for two compounds, for a mixture of more than one compound the average molecular weight of the mixture can be found as discussed in the above solution, that is the summation of the product of the mole fraction and molecular weight of each compound.