Question

The number of moles of oxygen in 1 L of air containing 21% oxygen by volume, in standard conditions, is:
(A) 0.186 mol
(B) 0.21 mol
(C) 2.10 mol
(D) 0.0093 mol

Answer 1

Hint: The % by volume refers to the concentration expression equal to ratio of volume of solute to the volume of solution, multiplied by 100. Using this we will calculate the volume of oxygen present in 1 L of air and we know that volume of 1 mole of any gas is 22.4 L at STP. Then we can calculate the number of moles by taking the ratio of volume of oxygen to volume of 1 mole oxygen at STP.

Complete step by step answer:
-First of all let us see what this % by volume means.
% v/v is an expression of concentration. It is basically defined as the volume of the solute per 100 parts by volume of the solution. Mathematically the expression of % v/v can be written as:
$\mathrm{\%}volume={\displaystyle \frac{vol.solute}{vol.solution}}\times 100$

-The question gives us the value of % v/v of oxygen = 21%
Using the above given equation we can now calculate the volume of oxygen in the following manner:
Let the volume of solute be V litre and the volume of the solution (which is the volume of air) is given to be 1 L.
$21={\displaystyle \frac{V}{1}}\times 100$
$V={\displaystyle \frac{21}{100}}\times 1$
V = 0.21 litre = 210 $c{m}^3$ or ml

-We all know that at STP conditions the volume of I mole of any gas = 22.414 L
  = 22414 $c{m}^3$ or ml
So, the volume of 1mole of oxygen = 22.414 L

-We will now calculate the number of moles of oxygen in 1 L of air and containing 21% oxygen by volume:
 $n={\displaystyle \frac{Vol.}{Vol.(STP)}}$
$={\displaystyle \frac{0.21}{22.4}}$
= 0.0093 moles
So, we can now say the number of moles of oxygen in 1 L of air containing 21% oxygen by volume, in standard conditions, is 0.0093 mol.
So, the correct answer is “Option D”.

Note: The standard conditions refer to the STP conditions where the temperature is 273.15K and 1 atm pressure. At these STP conditions the volume of 1 mole of gas will always be 22.4 litre and the number of atoms or molecules in it will be $6.022\times {10}^{23}$ ($N_A$). The expression of moles will be:
$n={\displaystyle \frac{Wt.given}{Mol.wt.}}$
$={\displaystyle \frac{Vol.given}{Vol.(STP)}}$
$={\displaystyle \frac{No.}{N_A}}$