Find the ratio of molecular volume to the actual volume occupied by the oxygen gas at STP when the diameter of an oxygen molecule is 3Å.
(A) $2 \times {10^{-4}}$
(B) $1 \times {10^{-4}}$
(C) $1.5 \times {10^{-4}}$
(D) $4 \times {10^{-4}}$

Hint
The molecular volume of the gas is the volume occupied by a molecule of gas i.e. the volume is multiplied by the Avogadro’s number. Whereas at STP, the volume is occupied by one mole of the gas.

Complete step by step answer
Given, the diameter of an oxygen molecule is 3Å $ = 3 \times {10^{ - 10}}$ m
Thus, the radius of an oxygen molecule, ${\rm{r}} = \dfrac{{\rm{d}}}{2} = \dfrac{{3 \times {{10}^{ - 10}}}}{2} = 1.5 \times {10^{ - 10}}$ m
Now, the molecular volume of oxygen gas, ${{\rm{V}}_{\rm{m}}} = \dfrac{4}{3}{\rm{\pi }}{{\rm{r}}^3} \times {{\rm{N}}_{\rm{A}}}$
Here ${{\rm{N}}_{\rm{A}}}$ is the Avogadro’s number whose value is $6.023 \times {10^{23}}$.
${{\rm{V}}_{\rm{m}}} = \dfrac{4}{3} \times 3.14 \times {\left( {1.5 \times {{10}^{ - 10}}} \right)^3} \times 6.023 \times {10^{23}}$
$ = 85.1 \times {10^{ - 7}}{\rm{\;}}{{\rm{m}}^3} = 8.51{\rm{\;c}}{{\rm{m}}^3}$ … (1)
Now, the actual volume of oxygen gas occupied by 1 mole at STP is given by,
 ${{\rm{V}}_{\rm{a}}} = 22400{\rm{\;c}}{{\rm{m}}^3}$ … (2)
By dividing equations (1) and (2), we get
$\dfrac{{{{\rm{V}}_{\rm{m}}}}}{{{{\rm{V}}_{\rm{a}}}}} = \dfrac{{8.51}}{{22400}} = 3.799 \times {10^{ - 4}} \approx 4 \times {10^{ - 4}}$
Therefore, (D) $4 \times {10^{ - 4}}$ is the required solution.

Additional Information
One STP defines as at 273 K temperature and the standard pressure of 1 atm, one mole of gas occupies 22.4 L volume. Avogadro’s number indicates the number of atoms or the number of molecules in a mole of any substance. Thus, it is used to define the molecular weight of the gas.

Note
While dividing the molecular volume to the actual volume, the units should be in the same system i.e. either in the MKS system or CGS system. The conversion of units should be done carefully.