Question

If the components of air are ${N_2} - 78\% ,{O_2} - 21\% ,Ar - 0.9\% $and $C{O_2} - 0.1\% $ by volume, what would be the molecular weight of air?

Answer 1

Hint: The molecular weight of the substance is equal to the algebraic sum of the individual molecular weights of the components of that substance multiplying by their amount in the given substance.
For example a compound X has two components A and B and their amount in the compound is a and b respectively and their molecular weights are c and d respectively then the molecular weight of the Compound X will be: $a \times c + b \times d$

Complete step by step answer:
The two terms molecular weight and molar mass are used interchangeably .The distinction between these two terms is critical to determine. The term molar mass is used for referring to the specific molecule or the individual molecule.
The molecular mass is measured by mass spectrometry be it be for a small to a medium sized molecule. The molecular mass of the proteins can be determined by the mass spectrometry. Usually the molecular mass term is frequently used.
Here for the given sample of air the components of air are: ${N_2} - 78\% ,{O_2} - 21\% ,Ar - 0.9\% $ and $C{O_2} - 0.1\% $ by volume means such amounts are present in 100 grams sample of air.
The molecular weight of $_2$ is 28 g, $O_2$ is 32 g , Ar is 40 g and $CO_2$ is 44 g
Therefore the average molar mass will be:

$\dfrac{{28 \times 78 + 32 \times 21 + 40 \times 0.9 + 44 \times 0.1}}{{100}}$

$ = 28.93 \approx 29$

So, the molecular weight of air will be 29 grams.

Note: The molecular mass and the weight term are usually taken with reference of the carbon-12 and is used as a standard measure to use for the calculations, Number of moles is also used as a standard for the chemical calculations determination.