Question

The average molar mass of air becomes more in presence of which gas if present in air:
[A] ${{H}_{2}}$
[B] ${{O}_{2}}$
[C] ${{C}_{2}}{{H}_{6}}$
[D] $C{{H}_{4}}$

Answer 1

Hint: Firstly, try to calculate the molecular weight of each of the given options. Compare their molecular weights by that of air. If their molecular weight is greater than that of air, its presence will affect the molar mass of air. The average molar mass of air is 28.97 g/mol.

Complete answer:
We know that the air around us is a mixture of several gases like nitrogen, oxygen, argon, carbon dioxide and other noble gases.
The molar mass of any substance is the number of grams of the substance in its molecular form which contains Avogadro’s number atoms of the substance.
Let us try calculating the molar mass of air. We know that air consists of 78.09 % of nitrogen, 20.95% of oxygen, 0.933% argon, 0.03% of carbon dioxide. The other gases are present in very small amounts thus their volume fraction will not affect the molar mass to that extent therefore we will not consider them.
If we want to convert the percentages in volume fraction, we will get 0.7809 of nitrogen in its volumetric fraction and oxygen will be 0.2095, argon is 0.0093 and carbon dioxide is 0.0003. Multiplying these values by the molecular mass of nitrogen, oxygen, argon and carbon dioxide and adding them we get the molar mass of air to be around 28.96 g/mol.
Now, let us calculate the molar mass of the molecules given to us.
Firstly, we have ${{H}_{2}}$. Atomic number of hydrogen is 1. Therefore, its molecular mass is 2.
Next, we have ${{O}_{2}}$. Atomic number of oxygen is 8. Therefore the molecular mass of ${{O}_{2}}$ is 16.
Next, we have ${{C}_{2}}{{H}_{6}}$. Atomic number of carbon is 6 and its molar mass is 12 and that of hydrogen is 1. Therefore, molecular weight of ${{C}_{2}}{{H}_{6}}=\left( 12\times 2 \right)+\left( 1\times 6 \right)=30$
Lastly, we have $C{{H}_{4}}$. Its molecular weight will be 12+4 = 16.
As we can see the molecular mass of ${{C}_{2}}{{H}_{6}}$ is higher than that of air. Therefore, its presence in air will increase the molar mass of air.

Therefore, the correct answer is option [C] ${{C}_{2}}{{H}_{6}}$.

Note:
We have mentioned above that the molar mass of any substance is the number of grams of the substance in its molecular form which contains Avogadro’s number atoms of the substance. This is due to the fact that the mass of a substance is dependent on the molecular weight of the substance. It is defined by the number of protons and neutrons in the atom. Therefore, the molar mass of a substance like air is dependent on the sum of volume fractions of each of the molecular component times their individual molecular weights.