Question

If average molecular weight of air is 29, then assuming ${N_2}$ and ${O_2}$ gasses are there which options are correct regarding composition of air:
i).75 % by mass of ${N_2}$
ii).75 % by moles ${N_2}$
iii).72.41 % by mass of ${N_2}$

Answer 1

Hint – The average atomic mass of air can be found out by- $\dfrac{{x(mol.massof{N_2})}}{{100}} + \dfrac{{(100 - x)(mol.massof{O_2})}}{{100}}$ , here x is the percentage of nitrogen, so oxygen will be (100 – x) %. Put the above formula equal to 29 and find the answer.
Formula used - $\dfrac{{x(mol.massof{N_2})}}{{100}} + \dfrac{{(100 - x)(mol.massof{O_2})}}{{100}} = 29$

Complete answer:
We have been given that the average molecular weight of air is 29.
Also, ${N_2}$ and ${O_2}$ are there in the air, according to the question.
So, let us assume that there is x % of nitrogen and (100 – x) % of oxygen.
So, we can write that-
$\dfrac{{x(mol.massof{N_2})}}{{100}} + \dfrac{{(100 - x)(mol.massof{O_2})}}{{100}} = 29$
Putting the value of mol. mass of ${N_2}$ and ${O_2}$ in the above formula we get-
$\dfrac{{x(2 \times 14)}}{{100}} + \dfrac{{(100 - x)(2 \times 16)}}{{100}} = 29$
solving further we get-
$
  \dfrac{{x(2 \times 14)}}{{100}} + \dfrac{{(100 - x)(2 \times 16)}}{{100}} = 29 \\
   \Rightarrow 28x + 32(100 - x) = 29 \times 100 \\
   \Rightarrow 28x - 32x = 2900 - 3200 \\
   \Rightarrow - 4x = - 300 \\
   \Rightarrow x = \dfrac{{300}}{4} = 75 \\
$
So, 75 % by mass of ${N_2}$
Hence, the correct answer is option (i).

Note – Whenever such types of questions appear, then always write the things given in the question first and then as mentioned in the solution, we have been given that the air contains nitrogen and oxygen, so assume the percentage of nitrogen to be x % and oxygen will be (100 – x)% and then by using the above formula, we found out the value of x.