Question

The volume occupied by one molecule of water (density = 1g/cc) is:
(A) 18 cc
(B) 22400 cc
(C) 6.023$\times$10$^{-23}$cc
(D) 3.0$\times$10$^{-23}$cc

Answer 1

Hint:In the question we are given with the value of density. So, we can calculate the molar mass of water. As we know volume is related to the mass, and density. In this way you can calculate the volume occupied in consideration with the Avogadro’s number.

Complete step by step solution:
> First, let us know about the term cc. It means the cubic centimetre.
As mentioned, we are given with the density of water i.e. 1g/cc, or 1g/ cm$^{3}$.
> We know that the molar mass of water is 18g/ mol.
- The molar volume of water can be defined by the molar mass of water divided by the density of water.
Therefore, Molar volume = $\dfrac{Mass}{Density}$
Molar volume = 18 cm$^{3}$/ mol
> We have to calculate the volume occupied by one molecule of water, thus we will consider the value of Avogadro’s number i.e. 6.023$\times$10$^{-23}$.
> Thus, volume of one molecule of water will be found out by dividing the molar volume to that of Avogadro’s number.
So, the volume of 1 molecule of water = 18/(6.023$\times$10$^{-23}$)
Volume of 1 molecule of water = 3.0$\times$10$^{-23}$cc
In the last, we can say that the volume occupied by one molecule of water is 3.0 $\times$10$^{-23}$cc.
Hence, the correct option is (D).

Note: Don’t get confused why we considered the value of Avogadro’s number. It is considered as we have to calculate the for one molecule of water, the 18cc/ mol is the volume calculated for one mole of water molecules, known to be molar volume.